A) 2 sec A
B) 2 cosec A
C) 2 tan A
D) 2 cot A
Correct Answer: B
Solution :
Expression \[=\frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}\] \[=\frac{{{\sin }^{2}}A+{{(1+cosA)}^{2}}}{\operatorname{sinA}(1+cosA)}\] \[=\frac{{{\sin }^{2}}A+1+2\cos A+{{\cos }^{2}}A}{\sin A(1+cosA)}\] \[=\frac{{{\sin }^{2}}A+{{\cos }^{2}}A+1+2\cos A}{\sin A(1+cosA)}\] \[=\frac{2+2\cos A}{\sin A(1+cosA)}\] \[=\frac{2(1+cosA)}{\operatorname{sinA}(1+cosA)}=\frac{2}{\sin A}\] \[=2\cos ecA\]
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