A) \[2cosec\theta \]
B) \[2sec\theta \]
C) \[2sin\theta \]
D) \[2cos\theta \]
Correct Answer: A
Solution :
(a) \[\frac{\sin \theta }{1+\cos \theta }+\frac{\sin \theta }{1-\cos \theta }\] \[=\frac{\sin \theta \left( 1-\cos \theta \right)+\sin \theta \left( 1+\cos \theta \right)}{\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)}\] \[=\frac{\sin \theta -\sin \theta \cos \theta +\sin \theta +\sin \theta \cos \theta }{1-{{\cos }^{2}}\theta }\] \[=\frac{2\sin \theta }{{{\sin }^{2}}}=2\cos ec\theta \]
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