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 :

Given ƒ(x) = (sin(tan^–1x) + sin(cot^–1x))² – 1 <br><br> = (sin(tan^–1x) + sin(${\pi \over 2}$ - tan^–1x))² – 1 <br><br> = (sin(tan^–1x) + cos(tan^–1x))² – 1 <br><br>= sin²(tan^–1x) + cos²(tan^–1x) + 2sin(tan^–1x)cos(tan^–1x) + 1 <br><br>= 1 + sin(2tan^–1x) - 1 <br><br>= sin(2tan^–1x) <br><br>Also given $${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$$ <br><br> Integrating both sides we get <br><br>y = ${1 \over 2}$ sin^-1 (f(x)) + C <br><br>= ${1 \over 2}$ sin^-1 (sin(2tan^–1x)) + C <br><br>Given $y\left( {\sqrt 3 } \right) = {\pi \over 6}$ mean x = $\sqrt 3$ and y = ${\pi \over 6}$ <br><br>$\therefore$ ${\pi \over 6}$ = ${1 \over 2}$ sin^-1 (sin(2tan^–1$\sqrt 3$)) + C <br><br>$\Rightarrow$ ${\pi \over 6}$ = ${1 \over 2}$ sin^-1 (sin(2$\times$${\pi \over 3}$)) + C <br><br>$\Rightarrow$ ${\pi \over 6}$ = ${1 \over 2}$ sin^-1 (${{\sqrt 3 } \over 2}$) + C <br><br>$\Rightarrow$ ${\pi \over 6}$ = ${1 \over 2}$ $\times$ ${\pi \over 3}$ + C <br><br>$\Rightarrow$ C = 0 <br><br>Now y(${ - \sqrt 3 }$) means when x = ${ - \sqrt 3 }$ then find y. <br><br>y = ${1 \over 2}$ sin^-1 (sin(2tan^–1x)) <br><br>= ${1 \over 2}$ sin^-1 (sin(2tan^–1(${ - \sqrt 3 }$))) <br><br>= ${1 \over 2}$ sin^-1 (sin(-2tan^–1(${ \sqrt 3 }$))) <br><br>= ${1 \over 2}$ sin^-1 (sin(-2$\times$${\pi \over 3}$)) <br><br>= ${1 \over 2}$ sin^-1 (-sin(2$\times$${\pi \over 3}$)) <br><br>= ${1 \over 2}$ sin^-1 (-${{\sqrt 3 } \over 2}$) <br><br>= ${1 \over 2}$ $\times$ -sin^-1 (${{\sqrt 3 } \over 2}$) <br><br>= ${1 \over 2}$ $\times$ -${\pi \over 3}$ <br><br>= -${\pi \over 6}$