Let $f(x)=\left(\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right)^{2}-1,|x|>1$. If $\frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}\left(\sin ^{-1}(f(x))\right)$ and $y(\sqrt{3})=\frac{\pi}{6}$, then $y(-\sqrt{3})$ is equal to:
Correct Option: , 2
$\frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}\left(\sin ^{-1} f(x)\right)$
$2 y=\sin ^{-1} f(x)+\mathrm{C}=\sin ^{-1}\left(\sin \left(2 \tan ^{-1} x\right)\right)+\mathrm{C}$
$\Rightarrow 2\left(\frac{\pi}{6}\right)=\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)+C$
$\frac{\pi}{3}=\frac{\pi}{3}+C \quad \therefore \quad C=0$
for $x=-\sqrt{3}, 2 y=\sin ^{-1}\left(\sin \left(\frac{-2 \pi}{6}\right)\right)+0$
$\Rightarrow 2 y=\frac{-\pi}{3} \Rightarrow y=\frac{-\pi}{6}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.