Let ƒ(x) = (sin(tan–1x) + sin(cot–1x))2 – 1, |x| > 1. If ${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$ and $y\left( {\sqrt 3 } \right) = {\pi \over 6}$, then y(${ - \sqrt 3 }$) is equal to :

Question

Let ƒ(x) = (sin(tan^–1x) + sin(cot^–1x))² – 1, |x| > 1.
If ${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$ and $y\left( {\sqrt 3 } \right) = {\pi \over 6}$, then y(${ - \sqrt 3 }$) is equal to :

${{5\pi } \over 6}$

$ - {\pi \over 6}$

${\pi \over 3}$

${{2\pi } \over 3}$

Solution

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