Prove the following

Question:

$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right),-1 Solution: Let$y=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right)$Putting$x^{2}=\cos 2 \theta \quad \therefore \theta=\frac{1}{2} \cos ^{-1} x^{2}y=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right)y=\tan ^{-1}\left(\frac{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}}\right)y=\tan \left(\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right)y=\tan ^{-1}\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right)y=\tan ^{-1}\left[\frac{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}\right]y=\tan ^{-1}\left[\frac{1+\tan \theta}{1-\tan \theta}\right]y=\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \cdot \tan \theta}\right]y=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\theta\right)\right]\Rightarrow y=\frac{\pi}{4}+\theta \Rightarrow y=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^{2}$Differentiating both sides w.r.t.$x\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}\right)+\frac{1}{2} \frac{d}{d x}\left(\cos ^{-1} x^{2}\right)=0+\frac{1}{2} \times \frac{-1}{\sqrt{1-x^{4}}} \cdot \frac{d}{d x}\left(x^{2}\right)=\frac{-1.2 x}{2 \sqrt{1-x^{4}}}=-\frac{x}{\sqrt{1-x^{4}}}$Thus,$\quad \frac{d y}{d x}=-\frac{x}{\sqrt{1-x^{4}}}\$