JEE MAIN - Mathematics (2019 - 12th April Morning Slot - No. 5)
For x $$ \in $$ (0, 3/2), let f(x) = $$\sqrt x $$ , g(x) = tan x and h(x) = $${{1 - {x^2}} \over {1 + {x^2}}}$$. If $$\phi $$ (x) = ((hof)og)(x), then $$\phi \left( {{\pi \over 3}} \right)$$
is equal to :
$$\tan {{7\pi } \over {12}}$$
$$\tan {{11\pi } \over {12}}$$
$$\tan {\pi \over {12}}$$
$$\tan {{5\pi } \over {12}}$$
Explanation
$$\phi \left( x \right) = \left( {\left( {hof} \right)og} \right)(x) = h\left( {\sqrt {\tan x} } \right)$$
$$ \Rightarrow \phi (x) = {{1 - \tan x} \over {1 + \tan x}} = \tan \left( {{\pi \over 4} - 4} \right)$$
$$ \therefore $$ $$\phi \left( {{\pi \over 3}} \right) = \tan \left( {{\pi \over 4} - {\pi \over 3}} \right)$$
$$ \Rightarrow \tan \left( { - {\pi \over {12}}} \right) = \tan {{11\pi } \over {12}}$$
$$ \Rightarrow \phi (x) = {{1 - \tan x} \over {1 + \tan x}} = \tan \left( {{\pi \over 4} - 4} \right)$$
$$ \therefore $$ $$\phi \left( {{\pi \over 3}} \right) = \tan \left( {{\pi \over 4} - {\pi \over 3}} \right)$$
$$ \Rightarrow \tan \left( { - {\pi \over {12}}} \right) = \tan {{11\pi } \over {12}}$$
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