Question

Find $\frac{dy}{dx}$; if $y={\tan }^{-1}\left(\frac{\sin x}{1+\cos x}\right)$

Answer 1

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

$\Rightarrow \tan y=\frac{\sin x}{1+\cos x}$

$\Rightarrow {\sec }^{2}y\frac{dy}{dx}=\frac{(1+\cos x)\frac{d}{dx}\left(\sin x\right)-\sin x\frac{d}{dx}(1+\cos x)}{{(1+\cos x)}^2}$

$\Rightarrow {\sec }^{2}y\frac{dy}{dx}=\frac{(1+\cos x)\left(\cos x\right)-\sin x(-\sin x)}{{(1+\cos x)}^2}$

$\Rightarrow {\sec }^{2}y\frac{dy}{dx}=\frac{\cos x+{\cos }^{2}x+{\sin }^{2}x}{{(1+\cos x)}^2}$

$\Rightarrow {\sec }^{2}y\frac{dy}{dx}=\frac{\cos x+1}{{(1+\cos x)}^2}$

$\Rightarrow {\sec }^{2}y\frac{dy}{dx}=\frac{1}{(1+\cos x)}$

$\Rightarrow (1+{\tan }^{2}y)\frac{dy}{dx}=\frac{1}{(1+\cos x)}$

$\Rightarrow (1+{\left(\frac{\sin x}{1+\cos x}\right)}^2)\frac{dy}{dx}=\frac{1}{(1+\cos x)}$

$\Rightarrow (1+\frac{{\sin }^{2}x}{{(1+\cos x)}^2})\frac{dy}{dx}=\frac{1}{(1+\cos x)}$

$\Rightarrow (1+\frac{{\sin }^{2}x}{{(1+\cos x)}^2})\frac{dy}{dx}=\frac{1}{(1+\cos x)}$

$\Rightarrow \left(\frac{1+{\cos }^{2}x+2\cos x+{\sin }^{2}x}{1+\cos x}\right)\frac{dy}{dx}=1$

$\Rightarrow \left(\frac{1+1+2\cos x}{1+\cos x}\right)\frac{dy}{dx}=1$

$\Rightarrow \left(\frac{2+2\cos x}{1+\cos x}\right)\frac{dy}{dx}=1$

$\Rightarrow \left(\frac{2(1+\cos x)}{1+\cos x}\right)\frac{dy}{dx}=1$

$\Rightarrow 2\frac{dy}{dx}=1$

$\therefore \frac{dy}{dx}=\frac{1}{2}$