A) \[2\sin \theta \]
B) \[2\cos \theta \]
C) \[2\text{cosec}\theta \]
D) \[2\text{sec }\theta \]
Correct Answer: C
Solution :
| [c] We have, \[\sqrt{\frac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\frac{\sec \theta +1}{\sec \theta -1}}\] |
| \[=\frac{(\sec \theta -1)+(\sec \theta +1)}{{}}=\frac{2\sec \theta }{{}}\] \[[{{\sec }^{2}}\theta -1={{\tan }^{2}}\theta ]\] |
| \[=\frac{2\sec \theta }{\tan \theta }=\frac{2}{\cos \theta }\times \frac{\cos \theta }{\sin \theta }=\frac{2}{\sin \theta }=2\cos ec\,\,\theta \] |
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