Question

If $y=\sin \left\{\tan {}^{-1}\sqrt{(}\frac{1-x}{1+x}\right)\}$ prove that $\frac{dy}{dx}=\frac{-x}{\sqrt{1-x^2}}$

Answer 1

$\begin{array}{c}y=\sin \left\{2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right\}\\ \\ Letx=\cos \theta \\ y=\sin \left\{2{\tan }^{-1}\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\right\}\\ =\sin \left\{2{\tan }^{-1}\sqrt{\frac{2{\sin }^2\frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}}\right\}\\ =\sin \left\{2{\tan }^{-1}\left(\tan \frac{\theta }{2}\right)\right\}\\ =\sin \left\{2\times \frac{\theta }{2}\right\}\\ y=\sqrt{1-{\cos }^2\theta }\\ y=\sqrt{1-x^2}\\ \frac{dy}{dx}=\frac{1}{2}\times \frac{-2x}{\sqrt{1-x^2}}\\ \frac{dy}{dx}=\frac{-x}{\sqrt{1-x^2}}\end{array}$

Proved.