Question

Find the derivative of ${\tan }^{-1}\left(\frac{\sin x}{1+\cos x}\right)$ with respect to ${\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)$.

Answer 1

${\tan }^{-1}\left(\frac{\sin x}{1+\cos x}\right)={\tan }^{-1}\left(\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2\cos \frac{x}{2}\cos \frac{x}{2}}\right)={\tan }^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2}$

Let $A={\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)={\tan }^{-1}\left(\cot \frac{x}{2}\right)={\tan }^{-1}\left(\tan (\frac{\pi }{2}-\frac{x}{2})\right)$

$B=\frac{\pi }{2}-\frac{x}{2}$

To find $\frac{dA}{dB}=\frac{\frac{dA}{dx}}{\frac{dB}{dx}}=\frac{\frac{1}{2}}{\frac{-1}{2}}=-1$