Question

${\tan }^{-1}\left[\frac{\cos x}{1+\sin x}\right]$ is equal to

• A. $\frac{\pi }{4}-\frac{x}{2}$, for $x\in (-\frac{\pi }{2},\frac{3\pi }{2})$
• B. $\frac{\pi }{4}-\frac{x}{2}$, for $x\in (-\frac{\pi }{2},\frac{\pi }{2})$
• C. $\frac{\pi }{4}-\frac{x}{2}$, for $x\in (\frac{3\pi }{2},\frac{5\pi }{2})$
• D. $\frac{\pi }{4}-\frac{x}{2}$, for $x\in (-\frac{3\pi }{2},-\frac{\pi }{2})$

Answer 1

Answer: (A), (B)

$\frac{\pi }{4}-\frac{x}{2}$, for $x\in (-\frac{\pi }{2},\frac{\pi }{2})$

Let ${\tan }^{-1}\left[\frac{\cos x}{1+\sin x}\right]={\tan }^{-1}A$....(i)

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

$=\left[\frac{\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}{1+\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}\right]$

$=\frac{1-{\tan }^2\frac{x}{2}}{{(1+{\tan }^2\frac{x}{2})}^2}$

$=\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}$

$=\tan (\frac{\pi }{4}-\frac{x}{2})$

From eq (i)

${\tan }^{-1}\left(\frac{\cos x}{1+\sin x}\right)={\tan }^{-1}\left[\tan (\frac{\pi }{4}-\frac{x}{2})\right]$

$\therefore {\tan }^{-1}\left(\tan \theta \right)=\theta$ if $\frac{-\pi }{2}<\theta <\frac{\pi }{2}$

$\therefore \frac{-\pi }{2}<\frac{\pi }{4}-\frac{x}{2}<\frac{\pi }{2}$

$\Rightarrow \frac{-3\pi }{4}<-\frac{x}{2}<\frac{\pi }{4}$

$\Rightarrow \frac{-\pi }{2}<x<\frac{3\pi }{2}$