Given ƒ(x) = (sin(tan^–1x) + sin(cot^–1x))² – 1

= (sin(tan^–1x) + sin($${\pi \over 2}$$ - tan^–1x))² – 1

= (sin(tan^–1x) + cos(tan^–1x))² – 1

= sin²(tan^–1x) + cos²(tan^–1x) + 2sin(tan^–1x)cos(tan^–1x) + 1

= 1 + sin(2tan^–1x) - 1

= sin(2tan^–1x)

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^–1x)) + 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^–1x))

= $${1 \over 2}$$ 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 ($${{\sqrt 3 } \over 2}$$)

= $${1 \over 2}$$ $$ \times $$ -$${\pi \over 3}$$

= -$${\pi \over 6}$$