NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions in Hindi in Hindi PDF Download
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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
$\mathbf{{\text{0}} \leqslant {\text{y}} \leqslant \pi}$
$ \mathbf{- \;\dfrac{\pi}{{\text{2}}} \leqslant {\text{y}} \leqslant \dfrac{\pi}{{\text{2}}}}$
$\mathbf{{{0 < y < \pi }}}$
$\mathbf{{\text{ - }}\dfrac{\pi}{{\text{2}}}{\text{ < y < }}\dfrac{\pi}{{\text{2}}}}$
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
$\mathbf{\pi}$
$\mathbf{{\text{ - }}\;\dfrac{\pi}{{\text{3}}}}$
$\mathbf{\dfrac{\pi}{3}}$
$\mathbf{\dfrac{{{{2\pi }}}}{3}}$
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
$\mathbf{\dfrac{{{{7\pi }}}}{{\text{6}}}}$
$\mathbf{\dfrac{{{{5\pi }}}}{{\text{6}}}}$
$\mathbf{\dfrac{\pi}{3}}$
$\mathbf{\dfrac{\pi}{{\text{6}}}}$
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
$\mathbf{\dfrac{1}{2}}$
$\mathbf{\dfrac{1}{3}}$
$\mathbf{\dfrac{1}{4}}$
$\mathbf{1}$
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
\[\mathbf{\pi}\]
\[\mathbf{{\text{ - }}\dfrac{\pi}{{\text{2}}}}\]
\[\mathbf{{\text{0}}}\]
\[\mathbf{{\text{2}}\sqrt 3} \]
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
$\mathbf{\dfrac{{\text{x}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}}$
$\mathbf{\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}}$
$\mathbf{\dfrac{{\text{1}}}{{\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} }}}$
$\mathbf{\dfrac{{\text{x}}}{{\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} }}}$
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
$\mathbf{{\text{0,}}\dfrac{{\text{1}}}{{\text{2}}}}$
$\mathbf{{\text{1,}}\dfrac{{\text{1}}}{{\text{2}}}}$
$\mathbf{{\text{0}}}$
$\mathbf{\dfrac{{\text{1}}}{{\text{2}}}}$
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
$\mathbf{\dfrac{\pi}{{\text{2}}}}$
$\mathbf{\dfrac{\pi}{{\text{3}}}}$
$\mathbf{\dfrac{\pi}{{\text{4}}}}$
$\mathbf{\dfrac{{{{3\pi }}}}{{\text{4}}}}$
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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