Question

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

• A. $\frac{\pi }{3}$
• B. $\frac{2\pi }{3}$
• C. $-\frac{\pi }{6}$
• D. $\frac{5\pi }{6}$

Answer 1

The correct option is C $-\frac{\pi }{6}$

Note: This is a BONUS question as multiple answers are possible.
$f\left(x\right)=\left[\sin \right({\tan }^{-1}x)+\sin ({\cot }^{-1}x){]}^2-1$
Put ${\tan }^{-1}x=\varphi$, where $\varphi \in (-\frac{\pi }{2},-\frac{\pi }{4})\bigcup (\frac{\pi }{4},\frac{\pi }{2})$

$f\left(x\right)={\left[\sin \right({\tan }^{-1}x)+\sin ({\cot }^{-1}x\left)\right]}^2-1$
$=[\sin \varphi +\cos \varphi {]}^2-1$
$=1+2\sin \varphi \cos \varphi -1$
$=\sin 2\varphi =\frac{2x}{1+x^2}$

It is given that $\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\left({\sin }^{-1}\right(f\left(x\right)\left)\right)$
$\Rightarrow \frac{dy}{dx}=\frac{-1}{1+x^2},\text{for}|x|>1$
$|x|>1\Rightarrow 1\Rightarrow x>1$ and $x<-1$

To get the value of $y(-\sqrt{3})$, we have to integrate the value of $\frac{dy}{dx}$.
To integrate the expression, the interval should be continuous. Therefore, we have to integrate the expression in both the intervals.
$\Rightarrow y=-{\tan }^{-1}x+C_1$, for $x>1$ and $y=-{\tan }^{-1}x+C_2$, for $x<-1$
For $x>1,C_1=\frac{\pi }{2}\because y\left(\sqrt{3}\right)=\frac{\pi }{6}$ is given.
But $C_2$ can't be determined as no other information is given for $x<-1$. Therefore, all the options can be true as $C_2$ can't be determined.