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# NCERT Solutions for Class 12 Maths Chapter 2 - In Hindi

Last updated date: 13th Apr 2024
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## NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions in Hindi in Hindi PDF Download

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Competitive Exams after 12th Science

## Access NCERT Solutions for Mathematics Chapter 2 – प्रतिलोम त्रिकोणमितीय फलन

### प्रश्नावली 2.1

निम्नलिखित के मुख्य मानों को ज्ञात कीजिए

1. $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}$

उत्तर: मान लीजिए कि, ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = y}}$

अतः ${\text{sin y = - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = - sin }}\left( {\dfrac{\pi}{{\text{6}}}} \right){\text{ = sin }}\left( {{\text{ - }}\dfrac{\pi}{{\text{6}}}} \right)$

हमें ज्ञात है कि ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]$ होता है और ${\text{sin }}\left( {{\text{ - }}\dfrac{\pi}{{\text{6}}}} \right){\text{ = - }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं |इसलिए, ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)$  का प्रमुख मान ${\text{ - }}\dfrac{\pi}{{\text{6}}}$ हैं

2. ${\mathbf{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {\text{3}} }}{{\text{2}}}} \right)}$

उत्तर:  मान लीजिए कि, ${\cos ^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt 3 }}{{\text{2}}}} \right){\text{ = y}}$

अतः ${\text{sin y = }}\dfrac{{\sqrt {\text{3}} }}{{\text{2}}}{\text{ = cos }}\left( {\dfrac{\pi}{{\text{6}}}} \right)$

हमें ज्ञात है कि ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]$ होता है और ${\text{sin }}\left( {{\text{ - }}\dfrac{\pi}{{\text{6}}}} \right){\text{ = - }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं |

इसलिए, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {\text{3}} }}{{\text{2}}}} \right)$  का प्रमुख मान $\dfrac{\pi}{{\text{6}}}$ हैं

3. $\mathbf{{\text{cose}}{{\text{c}}^{{\text{ - 1}}}}\left( 2 \right)}$

उत्तर:  मान लीजिए कि, ${\text{cosec}}{{\text{ }}^{{\text{ - 1}}}}\left( {\text{2}} \right){\text{ = y}}$

अतः ${\text{cosec y = 2 = cosec }}\left( {\dfrac{\pi}{{\text{6}}}} \right)$हमें ज्ञात है कि ${\text{cose}}{{\text{c}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]{\text{ - }}\left\{ {\text{0}} \right\}$ होता है और ${\text{cosec }}\left( {\dfrac{\pi}{{\text{6}}}} \right){\text{ = 2 }}$ हैं|

इसलिए, ${\text{cose}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{2 }}} \right)$  का प्रमुख मान $\dfrac{\pi}{{\text{6}}}$ हैं

4. $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\sqrt {\text{3}} } \right)}$

उत्तर:  मान लीजिए कि, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\sqrt {\text{3}} } \right){\text{ = y}}$

अतः ${\text{tan y = - }}\sqrt {\text{3}} {\text{ = - tan }}\left( {\dfrac{\pi}{{\text{3}}}} \right){\text{ = tan }}\left( {{\text{ - }}\dfrac{\pi}{{\text{3}}}} \right)$

हमें ज्ञात है कि ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]$ होता है और ${\text{tan }}\left( {{\text{ - }}\dfrac{\pi}{{\text{3}}}} \right){\text{ = - }}\sqrt {\text{3}}$ हैं

इसलिए, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\sqrt {\text{3}} } \right)$  का प्रमुख मान ${\text{ - }}\dfrac{\pi}{3}$ हैं

5. $\mathbf{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}$

उत्तर:  मान लीजिए कि, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = y}}$

अतः ${\text{cos y = - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = - cos }}\left( {\dfrac{\pi}{{\text{3}}}} \right){\text{ = cos }}\left( {{{\pi - }}\dfrac{\pi}{{\text{3}}}} \right){\text{ = cos }}\left( {\dfrac{{{{2 \pi }}}}{{\text{3}}}} \right)$

हमें ज्ञात है कि ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]$ होता है और ${\text{cos }}\left( {{\text{ - }}\dfrac{{{{2 \pi }}}}{{\text{3}}}} \right){\text{ = - }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं |

इसलिए, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)$  का प्रमुख मान $\dfrac{{{{2\pi }}}}{3}$ हैं

6. $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - 1 }}} \right)}$

उत्तर:  मान लीजिए कि, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - 1}}} \right){\text{ = y}}$

अतः ${\text{tan y = - 1 = - tan }}\left( {\dfrac{\pi}{4}} \right){\text{ = tan }}\left( {{\text{ - }}\dfrac{\pi}{4}} \right)$

हमें ज्ञात है कि ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]$ होता है और ${\text{tan }}\left( {{\text{ - }}\dfrac{\pi}{{\text{4}}}} \right){\text{ = - 1 }}$ हैं |

इसलिए, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - 1 }}} \right)$  का प्रमुख मान ${\text{ - }}\dfrac{\pi}{4}$ हैं

7. $\mathbf{{\text{se}}{{\text{c}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{2}}}{{\sqrt {\text{3}} }}} \right)}$

उत्तर:  मान लीजिए कि, ${\text{se}}{{\text{c}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{2}}}{{\sqrt {\text{3}} }}} \right){\text{ = y}}$

अतः ${\text{sec y = }}\dfrac{{\text{2}}}{{\sqrt {\text{3}} }}{\text{ = sec }}\left( {\dfrac{\pi}{{\text{6}}}} \right)$

हमें ज्ञात है कि ${\text{se}}{{\text{c}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]{\text{ - }}\left\{ {\dfrac{\pi}{{\text{2}}}} \right\}$ होता है और ${\text{sec }}\left( {\dfrac{\pi}{{\text{6}}}} \right){\text{ = }}\dfrac{{\text{2}}}{{\sqrt {\text{3}} }}$ हैं

इसलिए, ${\text{se}}{{\text{c}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{2}}}{{\sqrt {\text{3}} }}} \right)$  का प्रमुख मान $\dfrac{\pi}{{\text{6}}}$ हैं

8. $\mathbf{{\text{co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right)}$

उत्तर:  मान लीजिए कि, ${\text{co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right){\text{ = y}}$

अतः ${\text{sin y = }}\sqrt {\text{3}} {\text{ = cot }}\left( {\dfrac{\pi}{{\text{6}}}} \right)$

हमें ज्ञात है कि ${\text{co}}{{\text{t}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]$ होता है और ${\text{cot }}\left( {\dfrac{\pi}{{\text{6}}}} \right){\text{ = }}\sqrt {\text{3}}$ हैं |

इसलिए, ${\text{co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right)$  का प्रमुख मान $\dfrac{\pi}{{\text{6}}}$ हैं

9. $\mathbf{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}} \right)}$

उत्तर:  मान लीजिए कि, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}} \right){\text{ = y}}$

अतः ${\text{cos y = - }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}{\text{ = - cos }}\left( {\dfrac{\pi}{{\text{4}}}} \right){\text{ = cos }}\left( {{{\pi - }}\dfrac{\pi}{{\text{4}}}} \right){\text{ = cos }}\left( {\dfrac{{{{3\pi }}}}{{\text{4}}}} \right)$

हमें ज्ञात है कि ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{{0 , \pi }}} \right]$ होता है और ${\text{cos }}\left( {\dfrac{{{{3\pi }}}}{{\text{4}}}} \right){\text{ = - }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}$ हैं |

इसलिए, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}} \right)$  का प्रमुख मान $\dfrac{{{{3\pi }}}}{{\text{4}}}$ हैं

10. $\mathbf{{\text{cose}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{ - }}\sqrt {\text{2}} } \right)}$

उत्तर: मान लीजिए कि, ${\text{cose}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{ - }}\sqrt {\text{2}} } \right){\text{ = y}}$

अतः ${\text{cosec y = - }}\sqrt {\text{2}} {\text{ = - cosec }}\left( {\dfrac{\pi}{{\text{4}}}} \right){\text{ = cosec }}\left( {{\text{ - }}\dfrac{\pi}{{\text{4}}}} \right)$

हमें ज्ञात है कि ${\text{cose}}{{\text{c}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]{\text{ - }}\left\{ {\text{0}} \right\}$ होता है और ${\text{cosec }}\left( {{\text{ - }}\dfrac{\pi}{{\text{4}}}} \right){\text{ = - }}\sqrt {\text{2}}$ हैं |

इसलिए, ${\text{cose}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{ - }}\sqrt {\text{2}} } \right)$  का प्रमुख मान $\dfrac{\pi}{4}$ हैं

निम्नलिखित के मुख्य मानों को ज्ञात कीजिए

11. $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\text{1}} \right){\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ + si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}$

उत्तर:  मान लीजिए कि, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\text{1}} \right){\text{ = x }}$

अतः ${\text{tan x = 1 = tan }}\left( {\dfrac{\pi}{{\text{4}}}} \right)$

हमें ज्ञात है कि ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left( {{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right)$ होता है और ${\text{tan }}\left( {\dfrac{\pi}{{\text{4}}}} \right){\text{ = 1 }}$ हैं |

इसलिए, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\text{1}} \right){\text{ = }}\dfrac{\pi}{{\text{4}}}{\text{ }}$

मान लीजिए कि, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = y }}$

अतः ${\text{cos y = - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = - cos }}\left( {\dfrac{\pi}{{\text{3}}}} \right){\text{ = - cos }}\left( {{{\pi - }}\dfrac{\pi}{{\text{3}}}} \right){\text{ = cos }}\left( {\dfrac{{{{2\pi }}}}{{\text{3}}}} \right)$

हमें ज्ञात है कि ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]$ होता है और ${\text{cos }}\left( {\dfrac{{{{2\pi }}}}{{\text{3}}}} \right){\text{ = - }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं |

इसलिए, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = }}\dfrac{{{{2\pi }}}}{{\text{3}}}{\text{ }}$

मान लीजिए कि, ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = z }}$

अतः ${\text{sin z = - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = - sin }}\left( {\dfrac{\pi}{{\text{6}}}} \right){\text{ = sin }}\left( {{\text{ - }}\dfrac{\pi}{{\text{6}}}} \right)$

हमें ज्ञात है कि ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left( {{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right)$ होता है और ${\text{sin }}\left( {{\text{ - }}\dfrac{\pi}{{\text{6}}}} \right){\text{ = - }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं |

इसलिए, ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = - }}\dfrac{\pi}{{\text{6}}}{\text{ }}$

अब, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\text{1}} \right){\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ + si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)$ ${\text{ = }}\dfrac{\pi}{{\text{4}}}{\text{ + }}\dfrac{{{{2\pi }}}}{{\text{3}}}{\text{ - }}\dfrac{\pi}{{\text{6}}}$

${\text{ = }}\dfrac{{{{3\pi + 8\pi - 2\pi }}}}{{{\text{12}}}}$

${\text{ = }}\dfrac{{{{9\pi}}}}{{{\text{12}}}}{\text{ = }}\dfrac{{{{3\pi}}}}{{\text{4}}}$

12. $\mathbf{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ + 2si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)}$

उत्तर:  मान लीजिए कि, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = x }}$

अतः ${\text{cos x = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = cos }}\left( {\dfrac{\pi}{{\text{3}}}} \right)$

हमें ज्ञात है कि ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]$ होता है और ${\text{cos }}\left( {\dfrac{\pi}{{\text{3}}}} \right){\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं

इसलिए, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = }}\dfrac{\pi}{{\text{3}}}{\text{ }}$

मान लीजिए कि, ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = y }}$

अतः ${\text{sin y = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = sin }}\left( {\dfrac{\pi}{{\text{6}}}} \right)$

हमें ज्ञात है कि ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left( {{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right)$ होता है और ${\text{sin }}\left( {\dfrac{\pi}{{\text{6}}}} \right){\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}$ हैं |

इसलिए, ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = }}\dfrac{\pi}{{\text{6}}}{\text{ }}$

अब, ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ + 2si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)$ ${\text{ = }}\dfrac{\pi}{{\text{3}}}{\text{ + 2 X }}\dfrac{\pi}{{\text{6}}}$

${\text{ = }}\dfrac{\pi}{{\text{3}}}{\text{ + }}\dfrac{\pi}{{\text{3}}}{\text{ = }}\dfrac{{{{2\pi }}}}{{\text{3}}}$

13.  यदि $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x = y}}}$, तो

1. $\mathbf{{\text{0}} \leqslant {\text{y}} \leqslant \pi}$

2. $\mathbf{- \;\dfrac{\pi}{{\text{2}}} \leqslant {\text{y}} \leqslant \dfrac{\pi}{{\text{2}}}}$

3. $\mathbf{{{0 < y < \pi }}}$

4. $\mathbf{{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{ < y < }}\dfrac{\pi}{{\text{2}}}}$

उत्तर:  हमें ज्ञात है कि ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left( {{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right)$ होता है

इसलिए, $- \;\dfrac{\pi}{{\text{2}}} \leqslant {\text{y}} \leqslant \dfrac{\pi}{{\text{2}}}$

अतः विकल्प (B ) सही उत्तर हैं |

14. $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right){\text{ - se}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{ - 2}}} \right)}$ का मान बराबर है

1. $\mathbf{\pi}$

2. $\mathbf{{\text{ - }}\;\dfrac{\pi}{{\text{3}}}}$

3. $\mathbf{\dfrac{\pi}{3}}$

4. $\mathbf{\dfrac{{{{2\pi }}}}{3}}$

उत्तर:  दिया गया है,  ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right){\text{ - se}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{ - 2}}} \right)$

मान लीजिए कि, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right){\text{ = x }}$

अतः ${\text{tan x = }}\sqrt {\text{3}} {\text{ = tan }}\left( {\dfrac{\pi}{{\text{3}}}} \right)$

हमें ज्ञात है कि ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]$ होता है और ${\text{tan }}\left( {\dfrac{\pi}{{\text{3}}}} \right){\text{ = }}\sqrt {\text{3}}$ हैं |

इसलिए, ${\tan ^{{\text{ - 1}}}}\left( {\sqrt 3 } \right){\text{ = }}\dfrac{\pi}{{\text{3}}}{\text{ }}$

मान लीजिए कि, ${\text{se}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{ - 2}}} \right){\text{ = y }}$

अतः ${\text{sin y = - 2 = - sec }}\left( {\dfrac{\pi}{{\text{3}}}} \right){\text{ = sec }}\left( {{{\pi - }}\dfrac{\pi}{{\text{3}}}} \right){\text{ = sec }}\left( {\dfrac{{{{2\pi }}}}{{\text{3}}}} \right)$

हमें ज्ञात है कि ${\text{sec}}{{\text{ }}^{{\text{ - 1}}}}$ कि प्रमुख शाखा का परिसर $\left[ {{0, \pi }} \right]{\text{ - }}\left\{ {\dfrac{\pi}{{\text{2}}}} \right\}$ होता है और ${\text{sec }}\left( {\dfrac{{{{2\pi }}}}{{\text{3}}}} \right){\text{ = - 2 }}$ हैं

इसलिए, ${\text{se}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{ - 2}}} \right){\text{ = }}\dfrac{{{{2\pi }}}}{{\text{3}}}{\text{ }}$

अब, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\text{3}} } \right){\text{ - se}}{{\text{c}}^{{\text{ - 1}}}}\left( {{\text{ - 2}}} \right)$${\text{ = }}\dfrac{\pi}{{\text{3}}}{\text{ - }}\dfrac{{{{2\pi }}}}{{\text{3}}}{\text{ = - }}\dfrac{\pi}{{\text{3}}} अत: विकल्प (B) सही हैं | ### प्रश्नावली 2.2 निम्नलिखित को सिद्ध कीजिए 1. \mathbf{{\text{3si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{3x - 4}}{{\text{x}}^{\text{3}}}} \right){\text{, x}}\; \in \;\;\left[ {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{,}}\dfrac{{\text{1}}}{{\text{2}}}} \right]} उत्तर: {{x = sin \theta }} \Rightarrow {\text{ si}}{{\text{n}}^{{\text{ - 1}}}}{{x = \theta }} {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{3x - 4}}{{\text{x}}^{\text{3}}}} \right){\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{3sin x - 4si}}{{\text{n}}^{\text{3}}}{\text{ x}}} \right) \Rightarrow {\text{ si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{{sin 3\theta }}} \right) \Rightarrow {\text{ 3}}\theta \Rightarrow {\text{ 3 si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} 2. \mathbf{{\text{3co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x = co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{4}}{{\text{x}}^{\text{3}}}{\text{ - 3x}}} \right){\text{, x}}\; \in \;\;\left[ {\dfrac{{\text{1}}}{{\text{2}}}{\text{, 1 }}} \right]} उत्तर: {{x = cos }}\theta \Rightarrow {\text{ co}}{{\text{s}}^{{\text{ - 1}}}}{{x = \theta }} {\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{4}}{{\text{x}}^{\text{3}}}{\text{ - 3x}}} \right){\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{4 co}}{{\text{s}}^{\text{3}}}{{ \theta - 3 cos \theta }}} \right) \Rightarrow {\text{ co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{{cos 3\theta }}} \right) \Rightarrow {\text{ 3}}\theta \Rightarrow {\text{ 3 co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}} 3. \mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{2}}}{{{\text{11}}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{7}}}{{{\text{24}}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}} उत्तर: \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{2}}}{{{\text{11}}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{7}}}{{{\text{24}}}} \Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\dfrac{{\text{2}}}{{{\text{11}}}}{\text{ + }}\dfrac{{\text{7}}}{{{\text{24}}}}}}{{{\text{1 - }}\dfrac{{\text{2}}}{{{\text{11}}}}{\text{ }}{\text{. }}\dfrac{{\text{7}}}{{{\text{24}}}}}} \Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\dfrac{{{\text{48 + 77}}}}{{{{11 \times 24 }}}}}}{{\dfrac{{{{11 \times 24 - 14}}}}{{{{11 \times24 }}}}}} \Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{48 + 77}}}}{{{\text{264 - 14}}}}$${\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{125}}}}{{{\text{250 }}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}$

4: $\mathbf{{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{7}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{31}}}}{{{\text{17}}}}}$

उत्तर: $\Rightarrow 2\;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{1}{2}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{1}{7}$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2}}{\text{.}}\dfrac{{\text{1}}}{{\text{2}}}}}{{{\text{1 - }}{{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)}^{\text{2}}}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{7}}}$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{1}{{\left( {\dfrac{3}{4}} \right)}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{7}}}$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{4}{3}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{7}}}$

$\Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\dfrac{4}{3}{\text{ + }}\dfrac{1}{7}}}{{{\text{1 - }}\dfrac{4}{3}{\text{ }}{\text{. }}\dfrac{1}{7}}}$

$\Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\dfrac{{{\text{28 + 3}}}}{{{\text{21 }}}}}}{{\dfrac{{{\text{21 - 4}}}}{{{\text{21 }}}}}}$

$\Rightarrow {\text{ ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{31}}}}{{17}}$

निम्नलिखित फलनों को सरलतम रूप में लिखिए

5: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} {\text{ - 1}}}}{{\text{x}}}{\text{, x}} \ne {\text{0 }}}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} {\text{ - 1}}}}{{\text{x}}}{\text{,x}} \ne {\text{0}}$

${{x = tan\theta }} \Rightarrow {{\theta = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

$\therefore {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} {\text{ - 1}}}}{{\text{x}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{\text{1 + ta}}{{\text{n}}^{\text{2}}}{{\theta }}} {\text{ - 1}}}}{{{{tan\theta }}}}$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{{(sec \theta - 1)}}}}{{{{tan\theta }}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{{(1 - cos\theta )}}}}{{{{sin\theta }}}}$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2si}}{{\text{n}}^{\text{2}}}\dfrac{{{\theta }}}{{\text{2}}}}}{{{\text{2sin}}\dfrac{{{\theta }}}{{\text{2}}}{\text{cos}}\dfrac{{{\theta }}}{{\text{2}}}}}$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\dfrac{{{\theta }}}{{\text{2}}}} \right){\text{ = }}\dfrac{{{\theta }}}{{\text{2}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

6: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }},\;\left| {{\text{ x }}} \right|{\text{ > 1}}}$

उत्तर: ${{x = cosec\theta }} \Rightarrow {{\theta = cose}}{{\text{c}}^{{\text{ - 1}}}}{\text{x}}$

\begin{align} {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\sqrt {{\text{cose}}{{\text{c}}^{\text{2}}}{{\theta - 1}}} }} \hfill \\ {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{{{cot\theta }}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}{{(tan\theta )}} \hfill \\ {{ = \theta = cose}}{{\text{c}}^{{\text{ - 1}}}}{\text{x = }}\dfrac{\pi}{{\text{2}}}{\text{ - se}}{{\text{c}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ \end{align}

7: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\dfrac{{{\text{1 - cosx}}}}{{{\text{1 + cosx}}}}} {{ , 0 < x < \pi }}}$

उत्तर: $\Rightarrow {\text{ ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\dfrac{{{\text{1 - cosx}}}}{{{\text{1 + cosx}}}}} =$${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{\text{2co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}}}} \right)$

$\Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{sin}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}}}} \right){\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\dfrac{{\text{x}}}{{\text{2}}}} \right)$

$\Rightarrow \dfrac{{\text{x}}}{{\text{2}}}$

8: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{cos x - sin x}}}}{{{\text{cos x + sin x}}}}} \right)}$

उत्तर:  $\Rightarrow \;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1 - }}\dfrac{{{\text{sin x}}}}{{{\text{cos x}}}}}}{{{\text{1 + }}\dfrac{{{\text{sin x}}}}{{{\text{cos x}}}}}}} \right)$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1 - tan x}}}}{{{\text{1 + tan x}}}}} \right)$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\text{1}} \right){\text{ - ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan x}}} \right)$

$\Rightarrow \dfrac{\pi}{{\text{4}}}{\text{ - x}}$

9: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} }}{\text{ , }}\left| {{\text{ x }}} \right|{\text{ < a}}}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} }}{\text{ , }}\left| {{\text{ x }}} \right|{\text{ < a}}$

$\Rightarrow {\text{ ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} }}$

$\Rightarrow {{x = a sin \theta }}$

$\Rightarrow \dfrac{{\text{x}}}{{\text{a}}}{{ = sin \theta = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{x}}}{{\text{a}}}} \right)$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} }}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{{a sin \theta }}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}{\text{si}}{{\text{n}}^{\text{2}}}{{\theta }}} }}} \right)$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{{a sin \theta }}}}{{{\text{a}}\sqrt {{\text{1 - si}}{{\text{n}}^{\text{2}}}{{\theta }}} }}} \right){\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{{a sin \theta }}}}{{{{a cos \theta }}}}$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{(tan \theta ) = \theta = si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{a}}}$

10:$\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3}}{{\text{a}}^{\text{2}}}{\text{x - }}{{\text{x}}^{\text{3}}}}}{{{{\text{a}}^{\text{3}}}{\text{ - 3a}}{{\text{x}}^{\text{2}}}}}} \right){\text{, a > 0 ; - }}\dfrac{{\text{a}}}{{\sqrt {\text{3}} }}{\text{ < x < }}\dfrac{{\text{a}}}{{\sqrt {\text{3}} }}}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3}}{{\text{a}}^{\text{2}}}{\text{x - }}{{\text{x}}^{\text{3}}}}}{{{{\text{a}}^{\text{3}}}{\text{ - 3a}}{{\text{x}}^{\text{2}}}}}} \right)$

$\Rightarrow {{x = a tan \theta }}$

$\Rightarrow \dfrac{{\text{x}}}{{\text{a}}}{{ = tan \theta }} \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{a}}}$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3}}{{\text{a}}^{\text{2}}}{\text{x - }}{{\text{x}}^{\text{3}}}}}{{{{\text{a}}^{\text{3}}}{\text{ - 3a}}{{\text{x}}^{\text{2}}}}}} \right){\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3}}{{\text{a}}^{\text{2}}}{{.a tan \theta - }}{{\text{a}}^{\text{3}}}{\text{ta}}{{\text{n}}^{\text{3}}}{{\theta }}}}{{{{\text{a}}^{\text{3}}}{\text{ - 3a}}{\text{.}}{{\text{a}}^{\text{2}}}{\text{ta}}{{\text{n}}^{\text{2}}}{{\theta }}}}} \right)$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3}}{{\text{a}}^{\text{3}}}{{tan \theta - }}{{\text{a}}^{\text{3}}}{\text{ta}}{{\text{n}}^{\text{3}}}{{\theta }}}}{{{{\text{a}}^{\text{3}}}{\text{ - 3}}{{\text{a}}^{\text{3}}}{\text{ta}}{{\text{n}}^{\text{2}}}{{\theta }}}}} \right)$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{{3 tan \theta - ta}}{{\text{n}}^{\text{3}}}{{\theta }}}}{{{\text{1 - 3ta}}{{\text{n}}^{\text{2}}}{{\theta }}}}} \right)$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{(tan 3\theta )}}$

$\Rightarrow {{3\theta }}$

$\Rightarrow {\text{3 ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{a}}}$

निम्नलिखित में से प्रत्येक का मान ज्ञात कीजिए

11:$\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{2 cos}}\left( {{\text{2 si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}} \right)} \right]}$

उत्तर:  ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{2 cos}}\left( {{\text{2 si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}} \right)} \right]$

$\Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = x}}$

$\Rightarrow {\text{sin x = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = sin }}\left( {\dfrac{\pi}{{\text{6}}}} \right)$

$\Rightarrow \therefore {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{\pi}{{\text{6}}}$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{2 cos}}\left( {{\text{2 si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}} \right)} \right]{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{2 cos}}\left( {{\text{2 }}\left( {{\text{x}}\dfrac{\pi}{{\text{6}}}} \right)} \right)} \right]$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{2 cos}}\left( {\dfrac{\pi}{{\text{3}}}} \right)} \right]{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{2 x}}\dfrac{{\text{1}}}{{\text{2}}}} \right]$

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{1 = }}\dfrac{\pi}{{\text{4}}}$

12: $\mathbf{{\text{cot}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{a + co}}{{\text{t}}^{{\text{ - 1}}}}{\text{a}}} \right)}$

उत्तर:  ${\text{cot}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{a + co}}{{\text{t}}^{{\text{ - 1}}}}{\text{a}}} \right)$

$\cot\frac{\pi}{2} \;\;\;\;[\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2} ]=0$

13: $\mathbf{{\text{tan}}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2x}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}{\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{1 - }}{{\text{y}}^{\text{2}}}}}{{{\text{1 + }}{{\text{y}}^{\text{2}}}}}} \right]{\text{ , }}\left| {{\text{ x }}} \right|{\text{ < 1 , y > 0 or xy < 1}}}$

उत्तर:  ${\text{tan}}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2x}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}{\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{1 - }}{{\text{y}}^{\text{2}}}}}{{{\text{1 + }}{{\text{y}}^{\text{2}}}}}} \right]$

${{x = tan \theta }}$

$\Rightarrow {{\theta = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2x}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}{\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{{2 tan \theta }}}}{{{\text{1 + ta}}{{\text{n}}^{\text{2}}}\theta }}} \right){\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{{sin 2\theta }}} \right)$

${{ = 2\theta = 2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

${\text{y = tan}}\emptyset$

$\Rightarrow \emptyset \;{\text{ = }}\;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y}}$

$\therefore \;\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{1 - }}{{\text{y}}^{\text{2}}}}}{{{\text{1 + }}{{\text{y}}^{\text{2}}}}}{\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1 - tan}}{\emptyset ^{\text{2}}}}}{{{\text{1 + tan}}{\emptyset ^{\text{2}}}}}} \right){\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos2}}\emptyset } \right){\text{ = 2}}\emptyset$

${\text{ = 2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y}}$

$\therefore \;\;{\text{tan}}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2x}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}{\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{1 - }}{{\text{y}}^{\text{2}}}}}{{{\text{1 + }}{{\text{y}}^{\text{2}}}}}} \right]$

${\text{ = tan}}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{2 ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + 2 ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y}}} \right]$

${\text{ = tan}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y}}} \right]$

${\text{ = tan}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x + y}}}}{{{\text{1 - xy}}}}} \right)} \right]$

$=\left( {\frac{{{{x + y}}}}{{{{1 - xy}}}}} \right)$

14: $\mathbf{{\text{sin }}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}{\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ = 1}}}$

उत्तर: ${\text{sin }}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}{\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ = 1}}$

\begin{align} \Rightarrow {\text{sin}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}} \right){\text{ cos}}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ + cos}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}} \right){\text{ sin}}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right) \hfill \\ {\text{[ sin (A + B) = sin A cos B + cos A sin B ]}} \hfill \\ \dfrac{{\text{1}}}{{\text{5}}}{\text{X x + cos}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}} \right){\text{sin}}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ = 1}} \hfill \\ \dfrac{{\text{x}}}{{\text{5}}}{\text{ + cos}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}} \right){\text{sin }}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right){{ = 1, \ldots \ldots \ldots }}{\text{.(1)}} \hfill \\ {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}{\text{ = y}} \hfill \\ {\text{sin y = }}\dfrac{{\text{1}}}{{\text{5}}} \Rightarrow {\text{cos y = }}\sqrt {{\text{ 1 - }}{{\left( {\dfrac{{\text{1}}}{{\text{5}}}} \right)}^{\text{2}}}} {\text{ = }}\dfrac{{{\text{2}}\sqrt {\text{6}} }}{{\text{5}}} \Rightarrow {\text{y = co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2}}\sqrt {\text{6}} }}{{\text{5}}}} \right) \hfill \\ \therefore \;\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}{\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2}}\sqrt {\text{6}} }}{{\text{5}}}} \right){{ \ldots \ldots \ldots }}{\text{.(2)}} \hfill \\ {\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x = z}} \hfill \\ {\text{cosz = x}} \Rightarrow {\text{sinz = }}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} \Rightarrow {\text{z = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} } \right) \hfill \\ \therefore \;\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} } \right) \hfill \\ \end{align}

समीकरण (1),(2) और (3)

$\dfrac{{\text{x}}}{{\text{5}}}{\text{ + cos }}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2}}\sqrt {\text{6}} }}{{\text{5}}}} \right)} \right){\text{sin }}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} } \right)} \right){\text{ = 1}}$

\begin{align} \Rightarrow \dfrac{{\text{x}}}{{\text{5}}}{\text{ + }}\dfrac{{{\text{2}}\sqrt {\text{6}} }}{{\text{5}}}{\text{.}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ = 1}} \hfill \\ \Rightarrow {\text{x + 2}}\sqrt {\text{6}} {\text{.}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ = 5}} \hfill \\ \Rightarrow {\text{2}}\sqrt {\text{6}} {\text{.}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ = 5 - x}} \hfill \\ \Rightarrow {\left( {{\text{2}}\sqrt {\text{6}} {\text{.}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} } \right)^{\text{2}}}{\text{ = }}{\left( {{\text{5 - x}}} \right)^{\text{2}}} \hfill \\ \Rightarrow \left( {\text{4}} \right)\left( {\text{6}} \right)\;\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right){\text{ = 25 + }}{{\text{x}}^{\text{2}}}{\text{ - 10x}} \hfill \\ \Rightarrow {\text{24 - 24}}{{\text{x}}^{\text{2}}}{\text{ = 25 + }}{{\text{x}}^{\text{2}}}{\text{ - 10x}} \hfill \\ \Rightarrow {\text{25}}{{\text{x}}^{\text{2}}}{\text{ - 10x + 1 = 0}} \hfill \\ \Rightarrow {\left( {{\text{5x - 1}}} \right)^{\text{2}}}{\text{ = 0}} \hfill \\ \Rightarrow {\text{5x - 1 = 0}} \hfill \\ \Rightarrow {\text{5x = 1}} \hfill \\ \Rightarrow {\text{x = }}\dfrac{{\text{1}}}{{\text{5}}} \hfill \\ \end{align}

15: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{1 - x}}}}{{{\text{x - 2}}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{x + 1}}}}{{{\text{x + 2}}}}{\text{ = }}\dfrac{\pi}{{\text{4}}}}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{1 - x}}}}{{{\text{x - 2}}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{x + 1}}}}{{{\text{x + 2}}}}{\text{ = }}\dfrac{\pi}{{\text{4}}}$

\begin{align} \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {\dfrac{{\dfrac{{{\text{x - 1}}}}{{{\text{x - 2}}}}{\text{ + }}\dfrac{{{\text{x + 1}}}}{{{\text{x + 2}}}}}}{{{\text{1 - }}\dfrac{{{\text{x - 1}}}}{{{\text{x - 2}}}}{\text{.}}\dfrac{{{\text{x + 1}}}}{{{\text{x + 2}}}}}}} \right]{\text{ = }}\dfrac{\pi}{{\text{4}}} \hfill \\ \Rightarrow \left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{x + y}}}}{{{\text{1 - xy}}}}} \right]{\text{ = }}\dfrac{\pi}{{\text{4}}} \hfill \\ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {\dfrac{{\left( {{\text{x - 1}}} \right)\left( {\;{\text{x + 2}}} \right)\left( {{\text{x + 1}}} \right)\left( {{\text{x - 2}}} \right)}}{{\left( {\;{\text{x + 2}}} \right)\left( {{\text{x - 2}}} \right)\left( {{\text{x - 1}}} \right)\left( {{\text{x + 1}}} \right)}}} \right]\;{\text{ = }}\dfrac{\pi}{{\text{4}}} \hfill \\ \end{align}

\begin{align} \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {\dfrac{{\;{{\text{x}}^{\text{2}}}{\text{ + x - 2}}\;{\text{ + }}{{\text{x}}^{\text{2}}}{\text{ - x - 2}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 4 - }}{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} \right]{\text{ = }}\dfrac{\pi}{{\text{4}}} \hfill \\ \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {\dfrac{{{\text{2}}{{\text{x}}^{\text{2}}}{\text{ - 4}}}}{{{\text{ - 3}}}}} \right]{\text{ = }}\dfrac{\pi}{{\text{4}}} \hfill \\ \end{align}

$\Rightarrow {\text{tan}}\left[ {{\text{tan}}\dfrac{{{\text{4 - 2}}{{\text{x}}^{\text{2}}}}}{{\text{3}}}} \right]{\text{ = tan}}\dfrac{\pi}{{\text{4}}}$

\begin{align} \Rightarrow \dfrac{{{\text{4 - 2}}{{\text{x}}^{\text{2}}}}}{{\text{3}}}{\text{ = 1}} \hfill \\ \Rightarrow {\text{4 - 2}}{{\text{x}}^{\text{2}}}{\text{ = 3}} \hfill \\ \Rightarrow {\text{2}}{{\text{x}}^{\text{2}}}{\text{ = 4 - 3 = 1}} \hfill \\ \Rightarrow {{x = \pm }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }} \hfill \\ \end{align}

16 से 18 में प्रत्येक व्यंजक का मान ज्ञात कीजिए

16: $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin}}\dfrac{{{{2\pi }}}}{{\text{3}}}} \right)}$

उत्तर:   ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin}}\dfrac{{{{2\pi }}}}{{\text{3}}}} \right)$

\begin{align} \Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin x }}} \right){\text{ = x , x}} \in \left[ {\dfrac{{{{ - \pi }}}}{{\text{2}}},\dfrac{\pi}{{\text{2}}}} \right] \hfill \\ \Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin}}\dfrac{{{{2\pi }}}}{{\text{3}}}} \right){\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\left[ {\left( {{\text{sin}}\dfrac{{{{2\pi }}}}{{\text{3}}}} \right)} \right]{\text{ = }} \hfill \\ \Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin}}\left( {\dfrac{\pi}{{\text{3}}}} \right)} \right){\text{ , x}} \in \left[ {\dfrac{{{{ - \pi }}}}{{\text{2}}},\dfrac{\pi}{{\text{2}}}} \right] \hfill \\ \Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin}}\dfrac{{{{2\pi }}}}{{\text{3}}}} \right){\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{sin}}\left( {\dfrac{\pi}{{\text{3}}}} \right)} \right)\; = \;\dfrac{\pi}{{\text{3}}} \hfill \\ \end{align}

17: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\dfrac{{{{3\pi }}}}{{\text{4}}}} \right)}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\dfrac{{{{3\pi }}}}{{\text{4}}}} \right)$

\begin{align} \Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\dfrac{{{{3\pi }}}}{{\text{4}}}} \right){\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - tan}}\left( {{\text{ - }}\dfrac{{{{3\pi }}}}{{\text{4}}}} \right)} \right) \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - tan}}\left( {{{\pi - }}\dfrac{\pi}{{\text{4}}}} \right)} \right) \hfill \\ \end{align}

$\Rightarrow {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{ - tan}}\dfrac{\pi}{{\text{4}}}} \right]{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{tan}}\left( {{\text{ - }}\dfrac{\pi}{{\text{4}}}} \right)} \right]{\text{, x}} \in \left[ {{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]$

$\Rightarrow \therefore {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\left( {\dfrac{{{{3\pi }}}}{{\text{4}}}} \right)} \right){\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{tan}}\left( {{\text{ - }}\dfrac{\pi}{{\text{4}}}} \right)} \right]{\text{ = - }}\dfrac{\pi}{{\text{4}}}$

18: $\mathbf{{\text{tan}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{ + co}}{{\text{t}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{2}}}} \right)}$

उत्तर: $\Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{ = x}} \Rightarrow {\text{sin x = }}\dfrac{{\text{3}}}{{\text{5}}}$

\begin{align} \Rightarrow {\text{cos x = }}\sqrt {{\text{1 - si}}{{\text{n}}^{\text{2}}}{\text{x}}} {\text{ = }}\dfrac{{\text{4}}}{{\text{5}}}\;\; \Rightarrow {\text{sec x = }}\dfrac{{\text{5}}}{{\text{4}}} \hfill \\ \Rightarrow \therefore {\text{tan x = }}\sqrt {{\text{se}}{{\text{c}}^{\text{2}}}{\text{x - 1}}} {\text{ = }}\sqrt {\dfrac{{{\text{25}}}}{{{\text{16}}}}{\text{ - 1}}} {\text{ = }}\dfrac{{\text{3}}}{{\text{4}}} \hfill \\ \Rightarrow \therefore {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{4}}} \hfill \\ \Rightarrow \therefore {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{4}}}{\text{ }}............\left( {\text{1}} \right) \hfill \\ \Rightarrow {\text{co}}{{\text{t}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{2}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{2}}}{{\text{3}}}{\text{ }}...........\left( {\text{2}} \right) \hfill \\ \Rightarrow {\text{tan}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{ + co}}{{\text{t}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{2}}}} \right) \hfill \\ \Rightarrow {\text{tan}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\left( {\dfrac{{\text{3}}}{{\text{4}}}{\text{ + }}\dfrac{{\text{2}}}{{\text{3}}}} \right)}}{{{\text{1 - }}\dfrac{{\text{3}}}{{\text{4}}}{\text{ }}{\text{. }}\dfrac{{\text{2}}}{{\text{3}}}}}} \right)} \right) \hfill \\ \Rightarrow {\text{tan}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{9 + 8}}}}{{{\text{12 - 6}}}}} \right)} \right) \hfill \\ \Rightarrow {\text{tan}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{17}}}}{{\text{6}}}} \right)} \right){\text{ = }}\dfrac{{{\text{17}}}}{{\text{6}}} \hfill \\ \end{align}

19: $\mathbf{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right)}$ का मान

1. $\mathbf{\dfrac{{{{7\pi }}}}{{\text{6}}}}$

2. $\mathbf{\dfrac{{{{5\pi }}}}{{\text{6}}}}$

3. $\mathbf{\dfrac{\pi}{3}}$

4. $\mathbf{\dfrac{\pi}{{\text{6}}}}$

उत्तर: ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right)\;\;,\;\;\;x \in \left[ {0,\pi } \right]$

\begin{align} {\text{}} \Rightarrow {\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right){\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{ - 7\pi }}}}{{\text{6}}}} \right){\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\left( {{{2\pi - }}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right)} \right) \hfill \\ {\text{}} \Rightarrow {\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{5 \pi }}}}{{\text{6}}}} \right)\;\;\;\;{\text{,}}\dfrac{{{{5 \pi }}}}{{\text{6}}} \in \left[ {{0, \pi }} \right] \hfill \\ {\text{}} \Rightarrow \therefore \;\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right){\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{5 \pi }}}}{{\text{6}}}} \right){\text{ = }}\dfrac{{{{5 \pi }}}}{{\text{6}}} \hfill \\ \end{align}

(B) सही विकल्प है

20: $\mathbf{{\text{sin}}\left( {\dfrac{\pi}{{\text{3}}}{\text{ - si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)} \right)}$ का मान

1. $\mathbf{\dfrac{1}{2}}$

2. $\mathbf{\dfrac{1}{3}}$

3. $\mathbf{\dfrac{1}{4}}$

4. $\mathbf{1}$

उत्तर: ${\text{sin}}\left( {\dfrac{\pi}{{\text{3}}}{\text{ - si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)} \right)$

\begin{align} \Rightarrow {\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = x}} \hfill \\ \Rightarrow {\text{sin x = - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = - sin }}\dfrac{\pi}{{\text{6}}}{\text{ = sin }}\left( {{\text{ - }}\dfrac{\pi}{{\text{6}}}} \right) \hfill \\ \Rightarrow \therefore \,\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = - }}\dfrac{\pi}{{\text{6}}} \hfill \\ \Rightarrow \therefore \;\;{\text{sin}}\left( {\dfrac{\pi}{{\text{3}}}{\text{ - si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)} \right){\text{ = sin}}\left( {\dfrac{\pi}{{\text{3}}}{\text{ + }}\dfrac{\pi}{{\text{6}}}} \right) \hfill \\ {\text{sin}}\left( {\dfrac{\pi}{{\text{2}}}} \right){\text{ = 1}} \hfill \\ \end{align}

(D) सही विकल्प है

21: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{3}} {\text{ - co}}{{\text{t}}^{{\text{ - 1}}}}{\text{( - }}\sqrt {\text{3}} {\text{)}}}$ का मान

1. $\mathbf{\pi}$

2. $\mathbf{{\text{ - }}\dfrac{\pi}{{\text{2}}}}$

3. $\mathbf{{\text{0}}}$

4. $\mathbf{{\text{2}}\sqrt 3}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{3}} {\text{ - co}}{{\text{t}}^{{\text{ - 1}}}}{\text{( - }}\sqrt {\text{3}} {\text{)}}$

\begin{align} {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{( tan}}\dfrac{\pi}{{\text{3}}}{\text{) - co}}{{\text{t}}^{{\text{ - 1}}}}{\text{( - cot }}\dfrac{\pi}{{\text{6}}}{\text{)}} \hfill \\ \dfrac{\pi}{{\text{3}}}{\text{ - co}}{{\text{t}}^{{\text{ - 1}}}}\left[ {{\text{cot }}\left( {{{\pi - }}\dfrac{\pi}{{\text{6}}}} \right)} \right] \hfill \\ \dfrac{\pi}{{\text{3}}}{\text{ - co}}{{\text{t}}^{{\text{ - 1}}}}\left[ {{\text{cot }}\left( {\dfrac{{{{5\pi }}}}{{\text{6}}}} \right)} \right] \hfill \\ \dfrac{\pi}{{\text{3}}}{\text{ - }}\dfrac{{{{5\pi }}}}{{\text{6}}}{\text{ = }}\dfrac{{{{2\pi - 5\pi }}}}{{\text{6}}}{\text{ = - }}\dfrac{{{{3\pi }}}}{{\text{6}}}{\text{ = - }}\dfrac{\pi}{{\text{2}}} \hfill \\ \end{align}

(B) सही विकल्प है

### प्रश्नावली  A2

निम्नलिखित के मान ज्ञात कीजिए

1: $\mathbf{{\cos ^{{\text{ - 1}}}}\left( {{\text{cos }}\dfrac{{{{13\pi }}}}{{\text{6}}}} \right)}$

उत्तर: दिया गया है ${\cos ^{{\text{ - 1}}}}\left( {{\text{cos }}\dfrac{{{{13\pi }}}}{{\text{6}}}} \right)$

हमे ज्ञात है कि ${\text{x}} \in \left[ {{0, \pi }} \right]$ के लिए ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos x }}} \right){\text{ = x}}$

\begin{align} {\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\dfrac{{{{13\pi }}}}{{\text{6}}}} \right){\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos}}\;\left( {{{2\pi + }}\dfrac{\pi}{{\text{6}}}} \right)} \right) \hfill \\ {\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\left( {{\text{cos }}\left( {\dfrac{\pi}{{\text{6}}}} \right)} \right){\text{ = }}\dfrac{\pi}{{\text{6}}} \hfill \\ \end{align}

2: $\mathbf{{\tan ^{{\text{ - 1}}}}\left( {{\text{tan }}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right)}$

उत्तर: दिया गया है ${\tan ^{{\text{ - 1}}}}\left( {{\text{tan }}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right)$

हमे ज्ञात है कि ${\text{x}} \in \left[ {\dfrac{\pi}{{\text{2}}}{\text{, }}\dfrac{\pi}{{\text{2}}}} \right]$ के लिए ${\tan ^{{\text{ - 1}}}}\left( {\tan {\text{ x }}} \right){\text{ = x}}$

\begin{align} {\tan ^{{\text{ - 1}}}}\left( {\tan {\text{ }}\dfrac{{{{7\pi }}}}{{\text{6}}}} \right){\text{ = }}{\tan ^{{\text{ - 1}}}}\left( {\tan {\text{ }}\left( {{{\pi + }}\dfrac{\pi}{{\text{6}}}} \right)} \right) \hfill \\ {\text{ = }}{\tan ^{{\text{ - 1}}}}\left( {\tan {\text{ }}\left( {\dfrac{\pi}{{\text{6}}}} \right)} \right){\text{ = }}\dfrac{\pi}{{\text{6}}} \hfill \\ \end{align}

3: $\mathbf{{\text{2si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{24}}}}{{\text{7}}}}$

उत्तर: ${\text{2si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{ = 2ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\sqrt {{{\text{5}}^{\text{2}}}{\text{ - }}{{\text{3}}^{\text{2}}}} }}\;\;\;\;\;\;\;\;\;\;\quad \left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{p}}}{{\text{h}}}{\text{ = }}} \right.\left. {\begin{array}{*{20}{c}} {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{p}}}{{\sqrt {{{\text{h}}^{\text{2}}}{\text{ - }}{{\text{p}}^{\text{2}}}} }}} \end{array}} \right]{\text{ }}$

${\text{ = 2ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{4}}}$

\begin{align} {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2 X }}\dfrac{{\text{3}}}{{\text{4}}}}}{{{\text{1 - }}{{\left( {\dfrac{{\text{3}}}{{\text{4}}}} \right)}^{\text{2}}}}}\quad \;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2x}}}}{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}} \right] \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\dfrac{{\text{3}}}{{\text{2}}}}}{{\dfrac{{\text{7}}}{{{\text{16}}}}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{24}}}}{{\text{7}}} \hfill \\ \end{align}

4: $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{8}}}{{{\text{17}}}}{\text{ + si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{77}}}}{{{\text{36}}}}}$

उत्तर: ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{8}}}{{{\text{17}}}}{\text{ + si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{8}}}{{\sqrt {{\text{1}}{{\text{7}}^{\text{2}}}{\text{ - }}{{\text{8}}^{\text{2}}}} }}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\sqrt {{{\text{5}}^{\text{2}}}{\text{ - }}{{\text{3}}^{\text{2}}}} }}$

\begin{align} \left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{p}}}{{\text{h}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{p}}}{{\sqrt {{{\text{h}}^{\text{2}}}{\text{ - }}{{\text{p}}^{\text{2}}}} }}} \right] \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{8}}}{{{\text{15}}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{4}}} \hfill \\ \end{align}

${\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{8}}}{{{\text{15}}}}{\text{ + }}\dfrac{{\text{3}}}{{\text{4}}}}}{{{\text{1 - }}\dfrac{{\text{8}}}{{{\text{15}}}}{\text{ X }}\dfrac{{\text{3}}}{{\text{4}}}}}} \right)$

\begin{align} \left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x + y}}}}{{{\text{1 - x X y}}}}} \right)} \right] \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{77}}}}{{{\text{36}}}} \hfill \\ \end{align}

5: $\mathbf{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{4}}}{{\text{5}}}{\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{12}}}}{{{\text{13}}}}{\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{33}}}}{{{\text{65}}}}}$

उत्तर: ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{4}}}{{\text{5}}}{\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{12}}}}{{{\text{13}}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{{\text{5}}^{\text{2}}}{\text{ - }}{{\text{4}}^{\text{2}}}} }}{{\text{4}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{\text{1}}{{\text{3}}^{\text{2}}}{\text{ - 1}}{{\text{2}}^{\text{2}}}} }}{{{\text{12}}}}$

\begin{align} {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{4}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{5}}}{{{\text{12}}}} \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{5}}}{{{\text{12}}}}{\text{ + }}\dfrac{{\text{3}}}{{\text{4}}}}}{{{\text{1 - }}\dfrac{{\text{5}}}{{{\text{12}}}}{\text{ X }}\dfrac{{\text{3}}}{{\text{4}}}}}} \right) \hfill \\ \left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{y = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x + y}}}}{{{\text{1 - x X y}}}}} \right)} \right] \hfill \\ \end{align}

\begin{align} {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{56}}}}{{{\text{ 33 }}}} \hfill \\ {\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{33}}}}{{\sqrt {{\text{5}}{{\text{6}}^{\text{2}}}{\text{ + 3}}{{\text{3}}^{\text{2}}}} }} \hfill \\ {\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{33}}}}{{{\text{65}}}} \hfill \\ \end{align}

6: $\mathbf{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{12}}}}{{{\text{13}}}}{\text{ + si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{56}}}}{{{\text{65}}}}}$

उत्तर: ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{12}}}}{{{\text{13}}}}{\text{ + si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{\text{1}}{{\text{3}}^{\text{2}}}{\text{ - 1}}{{\text{2}}^{\text{2}}}} }}{{{\text{12}}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\sqrt {{{\text{5}}^{\text{2}}}{\text{ - }}{{\text{3}}^{\text{2}}}} }}$

\begin{align} {\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{56}}}}{{\sqrt {{\text{5}}{{\text{6}}^{\text{2}}}{\text{ + 3}}{{\text{3}}^{\text{2}}}} }} \hfill \\ {\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{56}}}}{{{\text{65}}}} \hfill \\ \end{align}

${\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{4}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{5}}}{{{\text{12}}}}$

\begin{align} {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{5}}}{{{\text{12}}}}{\text{ + }}\dfrac{{\text{3}}}{{\text{4}}}}}{{{\text{1 - }}\dfrac{{\text{5}}}{{{\text{12}}}}{\text{ X }}\dfrac{{\text{3}}}{{\text{4}}}}}} \right) \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{56}}}}{{{\text{33}}}}} \right) \hfill \\ \end{align}

7: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{63}}}}{{{\text{16}}}}{\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{5}}}{{{\text{13}}}}{\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}}$

उत्तर: ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{5}}}{{{\text{13}}}}{\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{3}}}{{\text{5}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{5}}}{{\sqrt {{\text{1}}{{\text{3}}^{\text{2}}}{\text{ - }}{{\text{5}}^{\text{2}}}} }}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{{\text{5}}^{\text{2}}}{\text{ - }}{{\text{3}}^{\text{2}}}} }}{{\text{3}}}{\text{ }}$

${\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{5}}}{{{\text{12}}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{4}}}{{\text{3}}}$

\begin{align} {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{5}}}{{{\text{12}}}}{\text{ + }}\dfrac{{\text{3}}}{{\text{4}}}}}{{{\text{1 - }}\dfrac{{\text{5}}}{{{\text{12}}}}{\text{ X }}\dfrac{{\text{3}}}{{\text{4}}}}}} \right) \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{63}}}}{{16}}} \right) \hfill \\ \end{align}

8:$\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{7}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{3}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{8}}}{\text{ = }}\dfrac{\pi}{{\text{4}}}}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{5}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{7}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{3}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{8}}}$

\begin{align} {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{1}}}{{\text{5}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{7}}}}}{{{\text{1 - }}\dfrac{{\text{1}}}{{\text{5}}}{\text{ X }}\dfrac{{\text{1}}}{{\text{7}}}}}} \right){\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{1}}}{{\text{8}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{3}}}}}{{{\text{1 - }}\dfrac{{\text{1}}}{{\text{8}}}{\text{ X }}\dfrac{{\text{1}}}{{\text{3}}}}}} \right) \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{12}}}}{{{\text{34}}}}{\text{ + ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{11}}}}{{{\text{23}}}} \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{{\text{12}}}}{{{\text{34}}}}{\text{ + }}\dfrac{{{\text{11}}}}{{{\text{23}}}}}}{{{\text{1 - }}\dfrac{{{\text{12}}}}{{{\text{34}}}}{\text{ X }}\dfrac{{{\text{11}}}}{{{\text{23}}}}}}} \right) \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{650}}}}{{{\text{650}}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1)}} \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\dfrac{\pi}{{\text{4}}}} \right){\text{ = }}\dfrac{\pi}{{\text{4}}} \hfill \\ \end{align}

सिद्ध कीजिए

9: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1 - x}}}}{{{\text{1 + x}}}}} \right){\text{, x}} \in {\text{[0, 1]}}}$

उत्तर: ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ X 2ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}}$

\begin{align} {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{1 + (}}\sqrt {\text{x}} {{\text{)}}^{\text{2}}}}}{{{\text{1 - (}}\sqrt {\text{x}} {{\text{)}}^{\text{2}}}}}\;\;\;\;\;\;\;\;\;\quad \left[ {{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}} \right] \hfill \\ {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1 - x}}}}{{{\text{1 + x}}}}} \right) \hfill \\ \end{align}

10: $\mathbf{{\text{co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1 + sin x}}} {\text{ + }}\sqrt {{\text{1 - sin x}}} }}{{\sqrt {{\text{1 + sinx }}} {\text{ - }}\sqrt {{\text{1 - sinx}}} }}} \right){\text{ = }}\dfrac{{\text{x}}}{{\text{2}}}{\text{, x}} \in \left( {{\text{0,}}\dfrac{\pi}{{\text{4}}}} \right)}$

उत्तर: ${\text{co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1 + sin x}}} {\text{ + }}\sqrt {{\text{1 - sin x}}} }}{{\sqrt {{\text{1 + sin x}}} {\text{ - }}\sqrt {{\text{1 - sin x}}} }}} \right)$

\begin{align} {\text{ = co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + 2cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{sin}}\dfrac{{\text{x}}}{{\text{2}}}} {\text{ + }}\sqrt {{\text{co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ - 2 cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{sin}}\dfrac{{\text{x}}}{{\text{2}}}} }}{{\sqrt {{\text{co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + 2cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{sin}}\dfrac{{\text{x}}}{{\text{2}}}} {\text{ - }}\sqrt {{\text{co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ - 2 cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{sin}}\dfrac{{\text{x}}}{{\text{2}}}} }}} \right) \hfill \\ {\text{ = co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\dfrac{{\left( {{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + sin}}\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{ + }}\left( {{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ - sin}}\dfrac{{\text{x}}}{{\text{2}}}} \right)}}{{\left( {{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + sin}}\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{ - }}\left( {{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ - sin}}\dfrac{{\text{x}}}{{\text{2}}}} \right)}}} \right) \hfill \\ {\text{ = co}}{{\text{t}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2cos}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{\text{2sin}}\dfrac{{\text{x}}}{{\text{2}}}}}} \right){\text{ = co}}{{\text{t}}^{{\text{ - 1}}}}\left( {{\text{cot}}\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{ = }}\dfrac{{\text{x}}}{{\text{2}}} \hfill \\ \end{align}

11: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1 + x}}} {\text{ - }}\sqrt {{\text{1 - x}}} }}{{\sqrt {{\text{1 + x}}} {\text{ + }}\sqrt {{\text{1 - x}}} }}} \right){\text{ = }}\dfrac{\pi}{{\text{4}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x , - }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }} \leqslant {\text{x}} \leqslant {\text{1 }}}$

उत्तर:    दिया गया है ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1 + x}}} {\text{ - }}\sqrt {{\text{1 - x}}} }}{{\sqrt {{\text{1 + x}}} {\text{ + }}\sqrt {{\text{1 - x}}} }}} \right)$

मान लेते है कि ${{x = cos t }}\; \Rightarrow {\text{t = co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}$

${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1 + x}}} {\text{ - }}\sqrt {{\text{1 - x}}} }}{{\sqrt {{\text{1 + x}}} {\text{ + }}\sqrt {{\text{1 - x}}} }}} \right){\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1 + cos t}}} {\text{ - }}\sqrt {{\text{1 - cos t}}} }}{{\sqrt {{\text{1 + cos t}}} {\text{ + }}\sqrt {{\text{1 - cos t}}} }}} \right)$

${\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{2 co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{t}}}{{\text{2}}}} {\text{ - }}\sqrt {{\text{2 si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{t}}}{{\text{2}}}} }}{{\sqrt {{\text{2co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{t}}}{{\text{2}}}} {\text{ + }}\sqrt {{\text{2si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{t}}}{{\text{2}}}} }}} \right){\text{ }}$

\begin{align} {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{cos}}\dfrac{{\text{t}}}{{\text{2}}}{\text{ - sin}}\dfrac{{\text{t}}}{{\text{2}}}}}{{\left. {{\text{cos}}\dfrac{{\text{t}}}{{\text{2}}}{\text{ + sin}}\dfrac{{\text{t}}}{{\text{2}}}} \right)}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1 - tan}}\dfrac{{\text{t}}}{{\text{2}}}}}{{{\text{1 + tan}}\dfrac{{\text{t}}}{{\text{2}}}}}} \right)} \right. \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{tan}}\dfrac{\pi}{{\text{4}}}{\text{ - tan}}\dfrac{{\text{t}}}{{\text{2}}}}}{{{\text{1 + tan}}\dfrac{\pi}{{\text{4}}}{\text{tan}}\dfrac{{\text{t}}}{{\text{2}}}}}} \right){\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{tan}}\left( {\dfrac{\pi}{{\text{4}}}{\text{ - }}\dfrac{{\text{t}}}{{\text{2}}}} \right)} \right){\text{ = }}\dfrac{\pi}{{\text{4}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ \end{align}

12: $\mathbf{\dfrac{{{{9\pi }}}}{{\text{8}}}{\text{ - }}\dfrac{{\text{9}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{3}}}{\text{ = }}\dfrac{{\text{9}}}{{\text{4}}}{\text{ si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2}}\sqrt {\text{2}} }}{{\text{3}}}}$

उत्तर:    $\dfrac{{{{9\pi }}}}{{\text{8}}}{\text{ - }}\dfrac{{\text{9}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{3}}}{\text{ = }}\dfrac{{\text{9}}}{{\text{4}}}\left( {\dfrac{\pi}{{\text{2}}}{\text{ - si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{3}}}} \right)$

\begin{align} {\text{ = }}\dfrac{{\text{9}}}{{\text{4}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{3}}}\;\;\;\;\;\;\;\quad \left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x = }}\dfrac{\pi}{{\text{2}}}} \right] \hfill \\ {\text{ = }}\dfrac{{\text{9}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {{{\text{3}}^{\text{2}}}{\text{ - }}{{\text{1}}^{\text{2}}}} }}{{\text{3}}} \hfill \\ {\text{ = }}\dfrac{{\text{9}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {\text{8}} }}{{\text{3}}}{\text{ = }}\dfrac{{\text{9}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2}}\sqrt {\text{2}} }}{{\text{3}}} \hfill \\ \end{align}

निम्नलिखित को सरल कीजिए

13: $\mathbf{{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(cos x) = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(2cosec x)}}}$

उत्तर:  दिया गया है ${\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(cos x) = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(2cosec x)}}$

\begin{align} {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2cos x}}}}{{{\text{1 - co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(2cosec x)}} \hfill \\ \left[ {{\text{2 ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{2x}}}}{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}} \right] \hfill \\ \end{align}

$\dfrac{{{\text{2cos x}}}}{{{\text{1 - co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{\text{ = 2 cosec x}} \Rightarrow \dfrac{{{\text{2 cos x}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{sin x}}}}$

\begin{align} \Rightarrow \dfrac{{{\text{cos x}}}}{{{\text{sin x}}}}{\text{ = 1}} \Rightarrow {\text{cot x = cot}}\dfrac{\pi}{{\text{4}}} \hfill \\ \Rightarrow {\text{x = }}\dfrac{\pi}{{\text{4}}} \hfill \\ \end{align}

14: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{1 - x}}}}{{{\text{1 + x}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x ,(x > 0)}}}$

उत्तर:  दिया गया है ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{1 - x}}}}{{{\text{1 + x}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

\begin{align} {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{1 - x}}}}{{{\text{1 + x}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{1 - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ \Rightarrow \dfrac{\pi}{{\text{4}}}{\text{ = }}\dfrac{{\text{3}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \Rightarrow \dfrac{\pi}{{\text{6}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ {\text{x = }}\dfrac{{\text{1}}}{{\sqrt {\text{3}} }} \hfill \\ \end{align}

15: $\mathbf{{\text{sin }}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ ,| x | < 1}}}$  बराबर होता है

1. $\mathbf{\dfrac{{\text{x}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}}$

2. $\mathbf{\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}}$

3. $\mathbf{\dfrac{{\text{1}}}{{\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} }}}$

4. $\mathbf{\dfrac{{\text{x}}}{{\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} }}}$

उत्तर: ${\text{sin}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ = }}$

\begin{align} {\text{sin}}{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} }}} \right)^{{\text{ - 1}}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\quad \left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{p}}}{{\text{b}}}{\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{p}}}{{\sqrt {{{\text{h}}^{\text{2}}}{\text{ + b}}} }}} \right] \hfill \\ {\text{sin}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ = }}\dfrac{{\text{x}}}{{\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} }} \hfill \\ \end{align}

(D) सही उत्तर है

16: $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1 - x) - 2 si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x = }}\dfrac{\pi}{{\text{2}}}}$ का मान बराबर है

1. $\mathbf{{\text{0,}}\dfrac{{\text{1}}}{{\text{2}}}}$

2. $\mathbf{{\text{1,}}\dfrac{{\text{1}}}{{\text{2}}}}$

3. $\mathbf{{\text{0}}}$

4. $\mathbf{\dfrac{{\text{1}}}{{\text{2}}}}$

उत्तर: मान लेते है कि ${{x = sin t }} \Rightarrow {\text{ t = si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

${\text{(1 - sin t) = sin}}\left( {\dfrac{\pi}{{\text{2}}}{\text{ + 2t}}} \right) \Rightarrow {\text{1 - sin t = cos 2t}}$

${\text{1 - sin t = 1 - 2si}}{{\text{n}}^{\text{2}}}{\text{t}} \Rightarrow {\text{2}}{{\text{x}}^{\text{2}}}{\text{ - x = 0}} \Rightarrow {\text{x (2x - 1) = 0}}$

${{x = 0, }}\dfrac{{\text{1}}}{{\text{2}}}$

(A) सही उत्तर है

17: $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{x}}}{{\text{y}}}} \right){\text{ - ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{x - y}}}}{{{\text{x + y}}}}}$  का मान

1. $\mathbf{\dfrac{\pi}{{\text{2}}}}$

2. $\mathbf{\dfrac{\pi}{{\text{3}}}}$

3. $\mathbf{\dfrac{\pi}{{\text{4}}}}$

4. $\mathbf{\dfrac{{{{3\pi }}}}{{\text{4}}}}$

उत्तर:   (C) सही उत्तर है

${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{x}}}{{\text{y}}}} \right){\text{ - ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{x - y}}}}{{{\text{x + y}}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\dfrac{{\text{x}}}{{\text{y}}}{\text{ - }}\dfrac{{{\text{x - y}}}}{{{\text{x + y}}}}}}{{{\text{1 + }}\dfrac{{\text{x}}}{{\text{y}}}{\text{ X }}\dfrac{{{\text{x - y}}}}{{{\text{x + y}}}}}}} \right)$

\begin{align} {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x(x + y) - y(x - y)}}}}{{{\text{y(x + y) + x(x - y)}}}}} \right) \hfill \\ {\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\left. {{{\text{x}}^{\text{2}}}{\text{ + xy - xy + }}{{\text{y}}^{\text{2}}}} \right)}}{{{{\text{x}}^{\text{2}}}{\text{ + xy - xy + }}{{\text{y}}^{\text{2}}}}}} \right){\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\left. {{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}} \right)}}{{{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}}}} \right){\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1) = }}\dfrac{\pi}{{\text{4}}} \hfill \\ \end{align}

(C) सही उत्तर है

### NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions In Hindi

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