Question

Prove:

$\sqrt{\frac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\frac{\sec \theta +1}{\sec \theta -1}}=2cosec\theta$

Answer 1

$LHS=\sqrt{\frac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\frac{\sec \theta +1}{\sec \theta -1}}=\sqrt{\frac{\frac{1}{\cos \theta }-1}{\frac{1}{\cos \theta }+1}}+\sqrt{\frac{\frac{1}{\cos \theta }+1}{\frac{1}{\cos \theta }-1}}(\because \sec \theta =\frac{1}{\cos \theta })=\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}+\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\sqrt{\frac{1-\cos \theta }{1+\cos \theta }\times \frac{1-\cos \theta }{1-\cos \theta }}+\sqrt{\frac{1+\cos \theta }{1-\cos \theta }\times \frac{1+\cos \theta }{1+\cos \theta }}=\sqrt{\frac{{(1-\cos \theta )}^2}{1-{\cos }^2\theta }}+\sqrt{\frac{{(1+\cos \theta )}^2}{1-{\cos }^2\theta }}=\sqrt{\frac{{(1-\cos \theta )}^2}{{\sin }^2\theta }}+\sqrt{\frac{{(1+\cos \theta )}^2}{{\sin }^2\theta }}(\because {\sin }^2\theta =1-{\cos }^2\theta )=\frac{1-\cos \theta }{\sin \theta }+\frac{1+\cos \theta }{\sin \theta }=\frac{2}{\sin \theta }=2cosec\theta (\because cosec\theta =\frac{1}{\sin \theta })$