Question

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

Answer 1

In right angled $\Delta ABC$,

Given, $\tan C=\sqrt{3}$

We know that, $\tan \theta =\frac{oppositesideto\angle \theta }{adjacentsideto\angle \theta }$

$\therefore AB=\sqrt{3}$ and $AC=1$

From Pythagoras theorem,

$B{C}^2=A{B}^2+A{C}^2$

$B{C}^2=(\sqrt{3}{)}^2+1^2$

$B{C}^2=3+1=4$

$BC=2$

$\sin \theta =\frac{oppositesideto\angle \theta }{hypotenuse}$

$\cos \theta =\frac{adjacentsideto\angle \theta }{hypotenuse}$

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

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

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

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

Therefore,

$\sin B\cos C+\cos BsinC$

$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$

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

$=\frac{4}{4}$

$=1$