Question

If $\tan \Theta =\frac{b}{a}$, then $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=$

• A. $\frac{2\sin \Theta }{\sqrt{\sin 2\Theta }}$
• B. $\frac{2\cos \Theta }{\sqrt{\cos 2\Theta }}$
• C. $\frac{2\cos \Theta }{\sqrt{\sin 2\Theta }}$
• D. $\frac{2\sin \Theta }{\sqrt{\cos 2\Theta }}$

Answer 1

Answer: (B)

$\frac{2\cos \Theta }{\sqrt{\cos 2\Theta }}$

Putting $b=a\tan \theta$ and then $\tan \theta =\frac{\sin \theta }{\cos \theta }$
We get,
$=\sqrt{\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }}+\sqrt{\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }}$
$=\frac{2\cos \theta }{\sqrt{co{s}^2\theta -{\sin }^2\theta }}$
$=\frac{2\cos \theta }{\sqrt{\cos 2\theta }}$