3+cosA1-2sinA-3+1+sinA3-2cosA-3=

3+2cos A1-2 sinA-3+1+2sin A3-2 cos A-3 =1-2 sinA3+2cosA3+3-2cosA1+2sinA3 =1-4sin2A3+3-4cos2A33+2cosA31+2sinA3 =1-4sin2A3-1-4sin2A33+2cosA31+2sinA3 =0

3+cosA1-2sinA-3+1+sinA3-2cosA-3=

3+2cos A1-2 sinA-3+1+2sin A3-2 cos A-3 =1-2 sinA3+2cosA3+3-2cosA1+2sinA3 =1-4sin2A3+3-4cos2A33+2cosA31+2sinA3 =1-4sin2A3-1-4sin2A33+2cosA31+2sinA3 =0

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${(\frac{\sqrt{3} + cosA}{1 - 2 sinA})}^{- 3} + {(\frac{1 + sinA}{\sqrt{3} - 2 cosA})}^{- 3} =$

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-1

$\sqrt{3}$

answer is C.

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Detailed Solution

$\begin{array}{rcl} {(\frac{\sqrt{3} + 2 \cos A}{1 - 2 sinA})}^{- 3} + {(\frac{1 + 2 \sin A}{\sqrt{3} - 2 \cos A})}^{- 3} \\ = & {(\frac{1 - 2 sinA}{\sqrt{3} + 2 cosA})}^{3} + {(\frac{\sqrt{3} - 2 cosA}{1 + 2 sinA})}^{3} \\ = & \frac{{(1 - 4 \sin^{2} A)}^{3} + {(3 - 4 \cos^{2} A)}^{3}}{{(\sqrt{3} + 2 cosA)}^{3} {(1 + 2 sinA)}^{3}} \\ = & \frac{{(1 - 4 \sin^{2} A)}^{3} - {(1 - 4 \sin^{2} A)}^{3}}{{(\sqrt{3} + 2 cosA)}^{3} {(1 + 2 sinA)}^{3}} \\ = & 0 \end{array}$