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