Question

Let $y={\tan }^{-1}\frac{1-\cos 2x}{\sin 2x}$, then prove that $\frac{dy}{dx}=-1$.

Answer 1

Given,

$y={\tan }^{-1}\frac{1-\cos 2x}{\sin 2x}$

or, $y={\tan }^{-1}\frac{2co{s}^{2}x}{2\sin x.\sin x}$

or, $y={\tan }^{-1}\frac{\cos x}{\sin x}$

or, $y={\tan }^{-1}\tan (\frac{\pi }{2}-x)$

or, $y=\frac{\pi }{2}-x$

Now differentiating both sides with respect to $x$ we get,

$\frac{dy}{dx}=-1$.