JEE MAIN - Mathematics (2019 - 12th April Evening Slot - No. 13)
The derivative of $${\tan ^{ - 1}}\left( {{{\sin x - \cos x} \over {\sin x + \cos x}}} \right)$$, with respect to $${x \over 2}$$
, where $$\left( {x \in \left( {0,{\pi \over 2}} \right)} \right)$$ is :
1
2
$${2 \over 3}$$
$${1 \over 2}$$
Explanation
$$y = {\tan ^{ - 1}}\left( {{{\tan x - 1} \over {\tan x + 1}}} \right)$$
$$ \Rightarrow $$ $$ - {\tan ^{ - 1}}\left( {{{1 - \tan x} \over {1 + \tan x}}} \right)$$
$$ \Rightarrow $$ $$ - {\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} - x} \right)} \right)$$
$$ \Rightarrow $$ $$ - \left( {{\pi \over 4} - x} \right)$$
$$ \Rightarrow $$ $${{dy} \over {dx}} = 1$$
Now if differentiation of $${x \over 2}$$ w.r.t is $${1 \over 2}$$
$$ \Rightarrow $$ So differentiation of y w.r.t $${x \over 2}$$ is $$1 \over {1 \over 2}$$ = 2
$$ \Rightarrow $$ $$ - {\tan ^{ - 1}}\left( {{{1 - \tan x} \over {1 + \tan x}}} \right)$$
$$ \Rightarrow $$ $$ - {\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} - x} \right)} \right)$$
$$ \Rightarrow $$ $$ - \left( {{\pi \over 4} - x} \right)$$
$$ \Rightarrow $$ $${{dy} \over {dx}} = 1$$
Now if differentiation of $${x \over 2}$$ w.r.t is $${1 \over 2}$$
$$ \Rightarrow $$ So differentiation of y w.r.t $${x \over 2}$$ is $$1 \over {1 \over 2}$$ = 2
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