Question

The derivative of ${\tan }^{-1}\left[\frac{\sin x}{1+\cos x}\right]$ with respect to ${\tan }^{-1}\left[\frac{\cos x}{1+\sin x}\right]$ is

• A. $2$
• B. $-1$
• C. $0$
• D. $-2$

Answer 1

Answer: (B)

$-1$

We have,

$\begin{array}{c}{y}_1={\tan }^{-1}\left(\frac{\sin x}{1+\cos x}\right)\\ ={\tan }^{-1}\left(\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{1+2{\cos }^2\frac{x}{2}-1}\right)\\ ={\tan }^{-1}\left(\tan \frac{x}{2}\right)\\ y_1=\frac{x}{2}\\ \frac{d{y}_1}{dx}=\frac{1}{2}\\ y_2={\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)\\ \frac{d{y}_2}{dx}=\frac{1}{2\left(1+\sin x\right)}.-\left(1+\sin x\right)\\ =\frac{1}{2}\\ \frac{d{y}_1}{dx}.\frac{dx}{d{y}_2}=\frac{1}{2}\times -2\\ \frac{d{y}_1}{d{y}_2}=-1\\ Hence,\\ optionBiscorrectanswer.\end{array}$