Question

$ \tan\left(\frac{\pi}{4} +\frac{\theta}{2}\right) + \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right)$ is equal to

• A. $\sec \, \theta $
• B. $2 \sec \, \theta $
• C. $\sec \, \frac {\theta} {2} $
• D. $\sin \, \theta $
• E. $\cos \, \theta $

Answer 1

Answer: (B)

We have,
$ \tan \left(\frac{\pi}{4}+\right. \left.\frac{\theta}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right) $
$= \frac{\tan \frac{\pi}{4}+\tan \frac{\theta}{2}}{1-\tan \frac{\pi}{4} \tan \frac{\theta}{2}}+\frac{\tan \frac{\pi}{4}-\tan \frac{\theta}{2}}{1+\tan \frac{\pi}{4} \tan \frac{\theta}{2}} $
$= \frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}+\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}} $
$= \frac{\left(1+\tan \frac{\theta}{2}\right)^{2}+\left(1-\tan \frac{\theta}{2}\right)^{2}}{1-\tan ^{2} \frac{\theta}{2}} $
$=\frac{1+\tan ^{2} \frac{\theta}{2}+2 \tan \frac{\theta}{2}+1+\tan ^{2} \frac{\theta}{2}-2 \tan \frac{\theta}{2}}{1-\tan ^{2} \frac{\theta}{2}}$
$=2\left[\frac{1+\tan ^{2} \theta / 2}{1-\tan ^{2} \theta / 2}\right]$
$=\frac{2}{\cos \theta} \,\,\,\left[\because \cos 2 \,\theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right]$
$=2 \sec \,\theta$