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Question

Prove that $$\displaystyle\sqrt{\frac{\sec \theta  - 1}{\sec \theta  + 1}} + \sqrt{\frac{\sec \theta  + 1}{\sec \theta  - 1}} = 2\text{cosec}\theta $$


Solution

$$\sqrt{\dfrac{\sec \theta-1}{\sec \theta+1}}+$$$$\sqrt{\dfrac{\sec \theta+1}{\sec \theta-1}}$$

$$=\dfrac{(\sec\theta-1)+(\sec\theta+1)}{\sqrt{\sec^2\theta-1}}$$

$$=\dfrac{2\sec \theta}{\tan \theta}=2\mbox{cosec}\theta$$

Mathematics

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