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