A) \[2a\]
B) \[3a\]
C) \[0\]
D) \[2ab\]
Correct Answer: A
Solution :
We have, \[sin\,\theta +cos\,\theta =a\] and \[sec\,\theta +cosec\,\theta =b\] Now, \[b({{a}^{2}}-1)=\left( \frac{1}{\cos \theta }+\frac{1}{\sin \theta } \right)\] \[({{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta -1)\] \[=\frac{\sin \theta +\cos \theta }{(\sin \theta \cos \theta )}\times (2\sin \theta \cos \theta )=2a\]You need to login to perform this action.
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