Question

Let $f\left(x\right)=\left(\sin \right({\tan }^{-1}x)+\sin {(co{t}^{-1}x))}^2-1$, where $\left|x\right|>1$.If

$\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\left(si{n}^{-1}f\right(x\left)\right)$

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

$\frac{–\pi }{6}$

Explanation for The correct option:

Finding the value of $y(-\sqrt{3})$

The given function,

$f\left(x\right)=\left(\sin \right({\tan }^{-1}x)+\sin {(co{t}^{-1}x))}^2-1....(i)$

$$\begin{gathered} \Rightarrow f\left(x\right)={\left(\sin \left({\tan }^{-1}x\right)+\sin \left(\frac{\pi }{2}-{\tan }^{-1}x\right)\right)}^2-1 \\ ={\left(\sin \left({\tan }^{-1}x\right)+\cos \left({\tan }^{-1}x\right)\right)}^2-1\left[\because \cos \theta =\sin \left(\frac{\mathrm{\pi }}{2}-\theta \right)\right] \\ ={\sin }^2\left({\tan }^{-1}x\right)+{\cos }^2\left({\tan }^{-1}x\right)+2\sin \left({\tan }^{-1}x\right)\cos \left({\tan }^{-1}x\right)-1 \\ =1+2\sin \left({\tan }^{-1}x\right)\cos \left({\tan }^{-1}x\right)-1\left[\because {\cos }^2\theta +{\sin }^2\theta =1\right] \\ =\sin (2{\tan }^{-1}x)...\left(ii\right)\left[\because 2\sin \theta \cos \theta =\sin 2\theta \right] \end{gathered}$$

As given,

$$\begin{gathered} \frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\left(si{n}^{-1}f\right(x\left)\right) \\ \Rightarrow 2\frac{dy}{dx}=\frac{d}{dx}\left(si{n}^{-1}f\right(x\left)\right) \end{gathered}$$

Integrating it w.r.t. $x$

$$\begin{gathered} \Rightarrow 2y\left(x\right)={\sin }^{-1}\left(\sin \left(2{\tan }^{-1}x\right)\right)+c....\left(iii\right) \\ \Rightarrow 2\times \frac{\pi }{6}={\sin }^{-1}\left(\sin \left(2{\tan }^{-1}\sqrt{3}\right)\right)+c\left[\because from\left(ii\right)f\left(x\right)=\sin \left(2{\tan }^{-1}x\right)andy\left(3\right)=\frac{\mathrm{\pi }}{6}isgiven\right] \\ \Rightarrow \frac{\pi }{3}={\sin }^{-1}\left(\sin 2\left(\frac{\pi }{3}\right)\right)+c\left[\because \tan \frac{\pi }{3}=\sqrt{3}\right] \\ \Rightarrow \frac{\pi }{3}={\sin }^{-1}\left(\sin \left(\mathrm{\pi }-\frac{\pi }{3}\right)\right)+c \\ \Rightarrow \frac{\pi }{3}=\frac{\pi }{3}+c \\ \Rightarrow c=0 \end{gathered}$$

Putting value of $c$ and $x=-\sqrt{3}$ into equation $\left(iii\right)$

$$\begin{gathered} \Rightarrow 2\times y\left(-\sqrt{3}\right)={\sin }^{-1}\left(\sin (2{\tan }^{-1}\left(-\sqrt{3}\right)\right) \\ ={\sin }^{-1}\left(\sin (2\times \frac{-\mathrm{\pi }}{3}\right) \\ =-\frac{\mathrm{\pi }}{3}\left[\because {\sin }^{-1}\left(\sin \theta \right)=\theta if\theta =\left[\frac{-\pi }{2},\frac{\pi }{2}\right]\right] \\ \Rightarrow 2y\left(-\sqrt{3}\right)=-\frac{\mathrm{\pi }}{3} \\ \Rightarrow y\left(-\sqrt{3}\right)=-\frac{\mathrm{\pi }}{6} \end{gathered}$$

Hence, the correct option is (C)