Question

$\frac{1+sin{600}^{\circ }-cos{600}^{\circ }}{1+sin{600}^{\circ }+cos{600}^{\circ }}$ = ?

• A.
• B. -
• C.
• D.

Answer 1

Answer: (A), (B)

The correct options are
A

B

-

Solution: Sin ${600}^{\circ }$ = $sin({360}^{\circ }+{240}^{\circ })$ = $sin{240}^{\circ }$

= sin $({180}^{\circ }+{60}^{\circ })$ = -sin ${60}^{\circ }$ = $-\frac{\sqrt{3}}{2}$

Cos${600}^{\circ }$ = cos$({360}^{\circ }+{240}^{\circ })$ = cos ${240}^{\circ }$

= cos$({180}^{\circ }+{60}^{\circ })$ = -cos ${60}^{\circ }$ = $-\frac{1}{2}$

$\Rightarrow$ Expression = $\frac{1+(-\frac{\sqrt{3}}{2})-(-\frac{1}{2})}{1+(-\frac{\sqrt{3}}{2})+(-\frac{1}{2})}$

= $\frac{1-\frac{\sqrt{3}}{2}+\frac{1}{2}}{1-\frac{\sqrt{3}}{2}-\frac{1}{2}}$ = $\frac{\frac{2-\sqrt{3}+1}{2}}{\frac{2-\sqrt{3}-1}{2}}$ = $\frac{3-\sqrt{3}}{1-\sqrt{3}}$

After rationalizing

= $\frac{3-\sqrt{3}}{1-\sqrt{3}}$ $\times$ $\frac{1+\sqrt{3}}{1+\sqrt{3}}$

= $\frac{3+3\sqrt{3}-\sqrt{3}-3}{1-3}$ = $\frac{2\sqrt{3}}{-2}$ = -$\sqrt{3}$