QJEEM20Mab226: JEE Main 2020 Mathematics - Class 12 JEEM20Mab226 Solution: Let ƒ(x) = (sin(tan –1 x) + sin(cot –1 x)) 2 – 1,

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

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

= - ${\pi \over 6}$