Question

If $\cos A=\cos B\cos C$ and $A+B+C=\pi$, then the value of $cotBcotC$ is

• A. $1$
• B. $2$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D

$\frac{1}{2}$

Finding the value of cot B cot C:

Given, $\begin{array}{rcl}A+B+C& =& \pi \\ A& =& \pi –(B+C)\end{array}$

Taking “$\cos$” on both sides,

$\begin{array}{rcl}cosA& =& cos[\pi –(B+C\left)\right]\\ cosA& =& -cos(B+C)\\ cosA& =& -[cosBcosC–sinBsinC]\\ cosBcosC& =& -cosBcosC+sinBsinC[\because giventhatcosA=cosBcosC]\\ \cos B\cos C+\cos B\cos C& =& \sin B\sin C\\ 2cosBcosC& =& sinBsinC\\ \frac{\cos B\cos C}{sinBsinC}& =& \frac{1}{2}\\ cotBcotC& =& 1/2\end{array}$

Hence, the correct answer is option (D)