Question
Without using a trigonometric table, prove that
$\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}-8 \sin ^2 30^{\circ}=0$

Answer

We have,
$\text { LHS }=\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}-8 \sin ^2 30^{\circ}=0 $
$=\frac{\cos \left(90^{\circ}-20^{\circ}\right)}{\sin 20^{\circ}}+\frac{\cos \left(90^{\circ}-31^{\circ}\right)}{\sin 31^{\circ}}-8 \times\left(\frac{1}{2}\right)^2$
$=\frac{\sin 20^{\circ}}{\sin 20^{\circ}}+\frac{\sin 31^{\circ}}{\sin 31^{\circ}}-8 \times \frac{1}{4}$
$= 1 + 1 - 2$
$= 2 -2$
$= 0$
$= \text{RHS}$
Hence proved.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free