A) \[2\,\text{cosec}\,A\]
B) \[2\sec A\]
C) \[2\,\sin A\]
D) \[2\cos A\]
Correct Answer: A
Solution :
[a] \[\frac{\sin A}{1+\cos A}+\frac{\sin A}{1-\cos A}\] \[=\frac{\sin A\,(1-\cos A)+\sin A\,(1+\cos A)}{(1+cos\,A)(1-cos\,A)}\] \[\sin A-\sin Acos\,A+\sin A\] \[=\frac{+\sin A\cos A}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1-{{\cos }^{2}}A}\] \[=\frac{2\sin A}{{{\sin }^{2}}A}=2\text{cosec}\,A\] |
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