A) \[1\]
B) \[-1\]
C) \[2\]
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
| [c]\[\frac{1}{1+\sin \theta }+\frac{1}{1-\sin \theta }=\frac{1-\sin \theta +1+\sin \theta }{(1+\sin \theta )(1-\sin \theta )}\] |
| \[[{{\sin }^{2}}+{{\cos }^{2}}\theta =1]\] |
| \[=\frac{2}{1-{{\sin }^{2}}\theta }=\frac{2}{{{\cos }^{2}}\theta }=2{{\sec }^{2}}\theta =k{{\sec }^{2}}\theta \] |
| On Comparing we get k=2 |
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