A) 1
B) \[\sqrt{3}\]
C) 0
D) \[-1\]
Correct Answer: C
Solution :
We have, \[{{\left( \frac{\sqrt{3}+2\cos A}{1-2\sin A} \right)}^{-3}}+{{\left( \frac{1+2\sin A}{\sqrt{3}-2\cos A} \right)}^{-3}}\] \[={{\left( \frac{1-2\sin A}{\sqrt{3}+2\cos A} \right)}^{3}}+{{\left( \frac{\sqrt{3}-2\cos A}{1+2\sin A} \right)}^{2}}\] \[=\frac{\begin{align} & {{[(1-2\sin A)\,(1+2\sin A)]}^{3}} \\ & \,\,+{{[(\sqrt{3}-2\cos A)\,(\sqrt{3}+2\cos A)]}^{3}} \\ \end{align}}{[(\sqrt{3}+2\cos A)\,\,{{(1+2\sin A)}^{3}}}=0\] \[=\frac{{{[1-4{{\sin }^{2}}A]}^{3}}+{{[3-4{{\cos }^{2}}A]}^{3}}}{{{(\sqrt{3}+2\cos A)}^{3}}\,{{(1+2\sin A)}^{3}}}=0\]
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