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