Question

Solve: ${\left(\frac{\sqrt{3}+2\cos A}{1-2\sin A}\right)}^{-3}+{\left(\frac{1+2\sin A}{\sqrt{3}-2\cos A}\right)}^{-3}$

• A. $1$
• B. $\sqrt{3}$
• C. $0$
• D. $-1$

Answer 1

The correct option is C $0$

${\left[\frac{\sqrt{3}+2\cos A}{1-2\sin A}\right]}^{-3}+{\left[\frac{1+2\sin A}{\sqrt{3}-2\cos A}\right]}^{-3}$

Put $A=90°$

$\Rightarrow {\left[\frac{\sqrt{3}+0}{1-2}\right]}^{-3}+{\left[\frac{1+2}{\sqrt{3}-0}\right]}^{-3}={\left(\frac{-1}{\sqrt{3}}\right)}^3+{\left(\frac{1}{\sqrt{3}}\right)}^3=0$