Question

In a triangle ABC,right angled at A, if tan $C=\sqrt{3}$, find the value of sin B cos C + cos B sin C.

Answer 1

tan C = $\sqrt{3}$

C = ${60}^{\circ }$

In triangle ABC ,

$\angle A+\angle B+\angle C={180}^{\circ }$

90 + $\angle B$ + 60 = 180

$\therefore \angle B=30$

sin B = $\frac{1}{2}$

cos C = $\frac{1}{2}$

cos B = $\frac{\sqrt{3}}{2}$

sin C = $\frac{\sqrt{3}}{2}$

$\therefore \frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}$

$=\frac{1}{4}+\frac{3}{4}$

$\Rightarrow \frac{4}{4}=1$