Question

If $A+B+C=\pi$ and $\cos A=\cos B.\cos C$ then show that $\cot B\cot C=\frac{1}{2}$.

Answer 1

Let us consider the problem.

Let $A+B+C={180}^0$ degrees

Hence

$A={180}^0-\left(B+C\right)$

Therefore

$\cos A=\cos B+\cos C$

Implies that

$\cos \left({180}^0-B-C\right)=\mathrm{cosBcosC}$

implies that

$-\cos \left(B+C\right)=\cos B.\cos C$

Implies that

$-\cos B\cos C+\sin B\sin C=\cos B.\cos C$

Implies that

$2\cos B\cos C=\sin B\sin C$

Implies that

Hence $\cot B\cot C=\frac{1}{2}$