Question

$\frac{d}{dx}\left\{{\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)\right\}\mathrm{equals}$

(a) 1/2
(b) − 1/2
(c) 1
(d) − 1

Answer 1

(b) − 1/2

$$\begin{gathered} \mathrm{Let}u={\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right) \\ \Rightarrow u={\tan }^{-1}\left(\frac{{\cos }^2{\displaystyle \frac{x}{2}}-{\sin }^2{\displaystyle \frac{x}{2}}}{{\cos }^2{\displaystyle \frac{x}{2}}+{\sin }^2{\displaystyle \frac{x}{2}}+2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}\right) \\ \Rightarrow u={\tan }^{-1}\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}^2} \\ \Rightarrow u={\tan }^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right) \\ \Rightarrow u={\tan }^{-1}\left[\frac{1-{\displaystyle \tan \frac{x}{2}}}{1+{\displaystyle \tan \frac{x}{2}}}\right] \\ \Rightarrow u={\tan }^{-1}\left[\frac{{\displaystyle \tan \frac{\mathrm{\pi }}{4}}-\tan {\displaystyle \frac{x}{2}}}{1+\tan \frac{\mathrm{\pi }}{4}\times \tan \frac{x}{2}}\right] \\ \Rightarrow u={\tan }^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{4}-\frac{x}{2}\right)\right] \\ \Rightarrow u=\frac{\mathrm{\pi }}{4}-\frac{x}{2} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{du}{dx}=0-\left(\frac{1}{2}\right) \\ \Rightarrow \frac{du}{dx}=-\frac{1}{2} \end{gathered}$$