Question

If $A+B+C=\pi$. The value $\frac{\cos A}{\sin B\sin C}+\frac{\cos B}{\sin C\sin A}+\frac{\cos C}{\sin A\sin B}$ is equal to

• A. $3$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A $3$

Consider the problem

$\begin{array}{c}LHS=\frac{\cos A}{\sin B\sin C}+\frac{\cos B}{\sin C\sin A}+\frac{\cos C}{\sin A\sin B}\\ =\frac{\cos A\sin A+\cos B\sin B+\cos C\sin C}{\sin A\sin B\sin C}\\ =\frac{2\cos A\sin A+2\cos B\sin B+2\cos C\sin C}{2\sin A\sin B\sin C}\left[OnDividingandmultiplyingby2\right]\\ =\frac{\sin 2A+\sin 2B+\sin 2C}{2\sin A\sin B\sin C}\end{array}$

Now, find the value of $sin2A+sin2B+sin2C$

$\begin{array}{c}A+B+C=\pi \\ \sin 2A+\sin 2B+\sin 2C\\ =2\sin \left(A+B\right)\cos \left(A-B\right)+2\sin C\cos C\\ =2\sin \left(\pi -C\right)\cos \left(A-B\right)+2\sin C\cos C\\ =2\sin C\cos \left(A-B\right)+2\sin C\cos C\\ =2\sin C\left(\cos \left(A-B\right)+\cos \left(\pi -\left(A+B\right)\right)\right)\\ =2\sin C\left(\cos \left(A-B\right)-\cos \left(A+B\right)\right)\\ =2\sin C\left(2\sin B\sin A\right)=4\sin A\sin B\sin C\end{array}$

$\begin{array}{c}LHS=\frac{4\sin A\times \sin B\times \sin C}{2\sin A\times \sin B\times \sin C}\\ =2\end{array}$

$=RHS$