Question

Prove that: $\sqrt{\frac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\frac{\sec \theta +1}{\sec \theta -1}}=2\csc \theta$.

Answer 1

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

$⟹\sqrt{\frac{(\sec \theta -1{)}^2}{(\sec \theta +1)(\sec \theta -1)}}$$+\sqrt{\frac{(\sec \theta +1{)}^2}{(\sec \theta -1)(\sec \theta +1)}}=2\csc \theta$

$⟹\sqrt{\frac{(\sec \theta -1{)}^2}{{\tan }^2\theta }}$$+\sqrt{\frac{(\sec \theta +1{)}^2}{{\tan }^2\theta }}=2\csc \theta$

$⟹\frac{2\sec \theta }{\tan \theta }$$=2\csc \theta$

Hence proved.