Question

$\frac{cos{70}^{\circ }}{sin{20}^{\circ }}+\frac{cos{59}^{\circ }}{sin{31}^{\circ }}-8si{n}^2{30}^{\circ }$ is equal to

• A. 1
• B. -1
• C. 0
• D. 2

Answer 1

The correct option is C

0

$\frac{cos{70}^{\circ }}{sin{20}^{\circ }}+\frac{cos{59}^{\circ }}{sin{31}^{\circ }}-8si{n}^2{30}^{\circ }$

$\because cos\theta =sin({90}^{\circ }-\theta ),sin{30}^{\circ }=\frac{1}{2}$

$\therefore cos70°=cos(90°-20°)=sin20°,\cos 59°=cos(90-31°)=sin31°$

$\Rightarrow \frac{sin{20}^{\circ }}{sin{20}^{\circ }}+\frac{sin{31}^{\circ }}{sin{31}^{\circ }}-8(\frac{1}{2}{)}^2$

$\Rightarrow 1+1-8\times \frac{1}{4}$
$\Rightarrow 2-2=0$