JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 10)
A possible value of $$\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)$$ is :
$$\sqrt 7 - 1$$
$${1 \over {\sqrt 7 }}$$
$$2\sqrt 2 - 1$$
$${1 \over {2\sqrt 2 }}$$
Explanation
$$\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)$$
$${\sin ^{ - 1}}\left( {{{\sqrt {63} } \over 8}} \right) = \theta $$ $$\sin \theta = {{\sqrt {63} } \over 8}$$
_24th_February_Evening_Shift_en_10_1.png)
$$\cos \theta = {1 \over 8}$$
$$2{\cos ^2}{\theta \over 2} - 1 = {1 \over 8}$$
$${\cos ^2}{\theta \over 2} = {9 \over {16}}$$
$$\cos {\theta \over 2} = {3 \over 4}$$
$${{1 - {{\tan }^2}{\theta \over 4}} \over {1 + {{\tan }^2}{\theta \over 4}}} = {3 \over 4}$$
$$\tan {\theta \over 4} = {1 \over {\sqrt 7 }}$$
$${\sin ^{ - 1}}\left( {{{\sqrt {63} } \over 8}} \right) = \theta $$ $$\sin \theta = {{\sqrt {63} } \over 8}$$
$$\cos \theta = {1 \over 8}$$
$$2{\cos ^2}{\theta \over 2} - 1 = {1 \over 8}$$
$${\cos ^2}{\theta \over 2} = {9 \over {16}}$$
$$\cos {\theta \over 2} = {3 \over 4}$$
$${{1 - {{\tan }^2}{\theta \over 4}} \over {1 + {{\tan }^2}{\theta \over 4}}} = {3 \over 4}$$
$$\tan {\theta \over 4} = {1 \over {\sqrt 7 }}$$
Comments (0)
