JEE MAIN - Mathematics (2022 - 25th June Evening Shift - No. 13)
The value of $${\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin \left( {{\pi \over 4}} \right)}}} \right)$$ is equal to :
$$ - {\pi \over 4}$$
$$ - {\pi \over 8}$$
$$ - {{5\pi } \over {12}}$$
$$ - {{4\pi } \over 9}$$
Explanation
$${\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin {\pi \over 4}}}} \right)$$
$$ = {\tan ^{ - 1}}\left( {{{{1 \over {\sqrt 2 }} - 1} \over {{1 \over {\sqrt 2 }}}}} \right)$$
$$ = {\tan ^{ - 1}}(1 - \sqrt 2 ) = - {\tan ^{ - 1}}(\sqrt 2 - 1)$$
$$ = - {\pi \over 8}$$
Comments (0)
