Question

If $0<x<\frac{\pi }{2}$, then $\tan (\frac{\pi }{4}+\frac{x}{2})+\tan (\frac{\pi }{4}-\frac{x}{2})$ is equal to

• A. $2\sec x$
• B. $2\cos x$
• C. $\sec x$
• D. $\cos x$
• E. $2\sin x$

Answer 1

Answer: (A)

$2\sec x$

Given, $0<x<\frac{\pi }{2}$, then
$\tan (\frac{\pi }{4}+\frac{x}{2})+\tan (\frac{\pi }{4}-\frac{x}{2})$
$=\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}+\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}$
$=\frac{{(1+\tan \frac{x}{2})}^2+{(1-\tan \frac{x}{2})}^2}{(1-{\tan }^2\frac{x}{2})}$
$=\frac{2(1+{\tan }^2\frac{x}{2})}{(1-{\tan }^2\frac{x}{2})}=\frac{2({\sin }^{2}x/2+{\cos }^{2}x/2)}{({\cos }^{2}x/2-{\sin }^{2}x/2)}$
$=\frac{2.1}{\cos x}=2\sec x$.