# Let f(x) =

Question:

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:

1. (1) $\frac{2 \pi}{3}$

2. (2) $-\frac{\pi}{6}$

3. (3) $\frac{5 \pi}{6}$

4. (4) $\frac{\pi}{3}$

Correct Option: , 2

Solution:

$\frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}\left(\sin ^{-1} f(x)\right)$

$2 y=\sin ^{-1} f(x)+\mathrm{C}=\sin ^{-1}\left(\sin \left(2 \tan ^{-1} x\right)\right)+\mathrm{C}$

$\Rightarrow 2\left(\frac{\pi}{6}\right)=\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)+C$

$\frac{\pi}{3}=\frac{\pi}{3}+C \quad \therefore \quad C=0$

for $x=-\sqrt{3}, 2 y=\sin ^{-1}\left(\sin \left(\frac{-2 \pi}{6}\right)\right)+0$

$\Rightarrow 2 y=\frac{-\pi}{3} \Rightarrow y=\frac{-\pi}{6}$