Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If x=$\tan {15^ \circ }$ ,y=${\text{cosec}}{75^ \circ }$ and z=$4\sin {18^ \circ }$ then,A) $x < y < z$ B) $y < z < x$ C) $z < x < y$ D) $x < z < y$

Last updated date: 25th Jun 2024
Total views: 414.9k
Views today: 10.14k
$\tan \left( {A + B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
$\sin \left( {A + B} \right) = \sin A\cos B + \operatorname{Cos} A\sin B$
And we know that $\sin {18^ \circ } = \dfrac{{\sqrt 5 - 1}}{4}$
We are given that x=$\tan {15^ \circ }$ -- (i)
y=${\text{cosec}}{75^ \circ }$ --- (ii)
and z=$4\sin {18^ \circ }$--- (iii)
We can write x=$\tan {15^ \circ }$$= \tan \left( {45 - 30} \right) And we know that \tan \left( {A + B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}} On substituting the given values we get, \Rightarrow x=\dfrac{{\tan {{45}^ \circ } - \tan {{30}^ \circ }}}{{1 + \tan {{45}^ \circ }\tan {{30}^ \circ }}} We know that \tan {45^ \circ } = 1 and \tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }} On substituting these values in the equation we get, \Rightarrow x=\dfrac{{1 - \dfrac{1}{{\sqrt 3 }}}}{{1 + \dfrac{1}{{\sqrt 3 }}}} \Rightarrow x=\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} On rationalization we get, \Rightarrow x= \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}} On solving the equation, we get- \Rightarrow x=\dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}} We know that \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} and{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab On using the formulae we get \Rightarrow x=\dfrac{{{{\left( {\sqrt 3 } \right)}^2} + 1 - 2 \times \sqrt 3 }}{{{{\left( {\sqrt 3 } \right)}^2} - 1}} On solving we get- \Rightarrow x= \dfrac{{3 + 1 - 2\sqrt 3 }}{{3 - 1}} \Rightarrow x=\dfrac{{4 - 2\sqrt 3 }}{2} = \dfrac{{2\left( {2 - \sqrt 3 } \right)}}{2} \Rightarrow x=2 - \sqrt 3 We know the value of \sqrt 3 = 1.732 then, \Rightarrow x=2 - 1.732 = 0.268 --- (iv) Now we know that {\text{cosec}}\theta {\text{ = }}\dfrac{1}{{\sin \theta }} and \sin \left( {A + B} \right) = \sin A\cos B + \operatorname{Cos} A\sin B, So we can write eq. (ii) as \Rightarrow y=${\text{cosec7}}{5^ \circ } = \dfrac{1}{{\sin {{75}^ \circ }}}$ \Rightarrow y=\dfrac{1}{{\sin \left( {40 + 30} \right)}} On using the formulae we get, \Rightarrow y=\dfrac{1}{{\sin 45\cos 30 + \cos 45\sin 30}} We know that \sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }} ,\cos 30 = \dfrac{{\sqrt 3 }}{2} ,\sin 30 = \dfrac{1}{2} and \cos 45 = \dfrac{1}{{\sqrt 2 }} On substituting these values we get, \Rightarrow y=\dfrac{1}{{\dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2}}} On simplifying we get, \Rightarrow y=\dfrac{1}{{\dfrac{{\sqrt 3 }}{{2\sqrt 2 }} + \dfrac{1}{{2\sqrt 2 }}}} On taking LCM, \Rightarrow y=\dfrac{1}{{\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}}} \Rightarrow y=\dfrac{{2\sqrt 2 }}{{\sqrt 3 + 1}} On rationalization, we get- \Rightarrow y=\dfrac{{2\sqrt 2 }}{{\sqrt 3 + 1}} \times \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}} = \dfrac{{2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{{{\left( {\sqrt 3 } \right)}^2} - 1}} \Rightarrow y=\dfrac{{2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{3 - 1}} On further simplifying we get- \Rightarrow y=\dfrac{{2\sqrt 6 - 2}}{2} = \dfrac{{2\left( {\sqrt 6 - \sqrt 2 } \right)}}{2} \Rightarrow y=\sqrt 6 - \sqrt 2 We know that \sqrt 2 = 1.414 and \sqrt 6 = 2.449. On substituting these values we get, \Rightarrow y=2.449 - 1.414 = 1.035 --- (v) Now, in eq. (ii) on putting the value \sin {18^ \circ } = \dfrac{{\sqrt 5 - 1}}{4}, we get z=4\sin {18^ \circ }=4 \times \dfrac{{\sqrt 5 - 1}}{4} = \sqrt 5 - 1 we know that \sqrt 5 = 2.23 On putting this value we get, z=2.23 - 1 = 1.23 -- (vi) From eq. (iv), (v), (vi), it is clear that x < y < z Hence, option A is correct. Note: We write \tan {15^ \circ }$$ = \tan \left( {45 - 30} \right)$ because we can easily calculate using the values of $\tan {45^ \circ }$ and $\tan {30^ \circ }$ . Similarly we can easily calculate the value of y using the values of $\sin {45^ \circ }$ and $\sin {30^ \circ }$ as we know the values of $\sin {45^ \circ }$ and $\sin {30^ \circ }$.