Question

In triangle $PQR$, right angled at $Q$, if $\tan P=\frac{1}{\sqrt{3}}$ find the value of $\cos P\cos R-\sin P\sin R$

Answer 1

$\tan P=\frac{1}{\sqrt{3}}$

$\Rightarrow \tan P=\tan 30°$

$\Rightarrow P=30°$

Now, $△PQR$ is a right angled at $Q,$ and $\angle P+\angle Q+\angle R=180°$

$\Rightarrow 30°+90°+\angle R=180°$

$\Rightarrow \angle R=60°$

Now, $\cos P\cos R-\sin P\sin R=\cos 30°\cos 60°-\sin 30°\sin 60°$

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

$=0.$

Hence, the answer is $0.$