Question : The value of the expression is:
$\frac{1+2\sin60^{\circ}\cos60^{\circ}}{\sin60^{\circ}+\cos60^{\circ}}+\frac{1-2\sin60^{\circ}\cos60^{\circ}}{\sin60^{\circ}-\cos60^{\circ}}$
Option 1: $2\sqrt{3}$
Option 2: $0$
Option 3: $\sqrt{3}$
Option 4: $2$
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Correct Answer: $\sqrt{3}$
Solution :
$\frac{1+2\sin60^{\circ}\cos60^{\circ}}{\sin60^{\circ}+\cos60^{\circ}}+\frac{1-2\sin60^{\circ}\cos60^{\circ}}{\sin60^{\circ}-\cos60^{\circ}}$
$=\frac{1+2×\frac{\sqrt3}{2}×\frac{1}{2}}{\frac{\sqrt3}{2}+\frac{1}{2}}+\frac{1-2×\frac{\sqrt3}{2}×\frac{1}{2}}{\frac{\sqrt3}{2}-\frac{1}{2}}$
$=\frac{1+\frac{\sqrt3}{2}}{\frac{\sqrt3+1}{2}}+\frac{1-\frac{\sqrt3}{2}}{\frac{\sqrt3-1}{2}}$
$=\frac{2+\sqrt3}{\sqrt3+1}+\frac{2-\sqrt3}{\sqrt3-1}$
Rationalize,
$=\frac{(2+\sqrt3)(\sqrt3-1)}{2}+\frac{(2-\sqrt3)(\sqrt3+1)}{2}$
$=\frac{2\sqrt3-2+3-\sqrt3}{2}+\frac{2\sqrt3+2-3-\sqrt3}{2}$
$=\frac{\sqrt3+1}{2}+\frac{\sqrt3-1}{2}$
$=\frac{2\sqrt3}{2}$
$=\sqrt3$
Hence, the correct answer is $\sqrt3$.
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