Question

Prove the identity

$\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}-\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\frac{2}{cot\theta }$

Answer 1

Prove the identity

Given data: $\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}-\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\frac{2}{cot\theta }$

Proof:

L. H. S

$$\begin{gathered} =\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}-\sqrt{\frac{1-\sin \theta }{1+\sin \theta }} \\ =\sqrt{\frac{1+\sin \theta }{1-\sin \theta }\times \frac{1-\sin \theta }{1-\sin \theta }}-\sqrt{\frac{1-\sin \theta }{1+\sin \theta }\times \frac{1+\sin \theta }{1+\sin \theta }} \\ =\sqrt{\frac{1-{\sin }^2\theta }{{(1-\sin \theta )}^2}}-\sqrt{\frac{1-{\sin }^2\theta }{{(1+\sin \theta )}^2}} \\ =\sqrt{\frac{{\cos }^2\theta }{{(1-\sin \theta )}^2}}-\sqrt{\frac{{\cos }^2\theta }{{(1+\sin \theta )}^2}} \\ =\frac{\cos \theta }{1-\sin \theta }-\frac{\cos \theta }{1+\sin \theta } \\ =\cos \theta \left[\frac{1}{1-\sin \theta }-\frac{1}{1+\sin \theta }\right] \\ =\cos \theta \left[\frac{1+\sin \theta -1+\sin \theta }{1-{\sin }^2\theta }\right] \\ =\cos \theta \times \frac{2\sin \theta }{{\cos }^2\theta } \\ =\frac{2\sin \theta }{\cos \theta } \\ =2\tan \theta \\ =\frac{2}{cot\theta }\left[\because \tan \theta =\frac{1}{cot\theta }\right] \end{gathered}$$

= R. H. S

Therefore, L. H. S = R. H. S

Hence, $\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}-\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\frac{2}{cot\theta }$ is proved.