Question

If y = $si{n}^{-}1\left(\frac{1}{\sqrt{1+x^2}}\right)+ta{n}^{-}1\left(\frac{\sqrt{1+x^2-1}}{x}\right),find\frac{dy}{dx}$

Answer 1

Let $x=\tan \theta$

$\Rightarrow y={\sin }^{-1}\left(\frac{1}{\sqrt{\sec 2\theta }}\right)+{\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)$

$\Rightarrow y={\sin }^1\left(\cos \theta \right)+{\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)$

As $1-\cos \theta =2{\sin }^2\theta /2$ & $\sin \theta =2\sin \theta /2\cos \theta /2$

$\Rightarrow y={\sin }^{-1}\left(\cos \theta \right)+{\tan }^{-1}\left(\tan \theta /2\right)$

$\Rightarrow y=\frac{\pi }{2}-\theta +\theta /2=\frac{\pi }{2}-\frac{\theta }{2}$

$\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}=(-\frac{1}{2})\left(\frac{1}{1+x^2}\right)=\frac{-1}{2(1+x^2)}$.