Question

In $\Delta$ $ABC$, the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively, then $\frac{bc{\sin }^{2}A}{\cos A+\cos B\cos C}$ is equal to?

• A. $b^2+c^2$
• B. $bc$
• C. $a^2$
• D. $a^2+bc$

Answer 1

The correct option is C $a^2$

$\frac{bc{\sin }^{2}A}{\cos A+\cos B\cos C}=\frac{bc{\sin }^{2}A}{\cos (\pi -(B+C\left)\right)+\cos B\cos C}=\frac{bc{\sin }^{2}A}{-\cos (B+C)+\cos B\cos C}=\frac{bc{\sin }^{2}A}{-\cos B\cos C+\sin B\sin C+\cos B\cos C}=\frac{bc{\sin }^{2}A}{\sin B\sin C}=\frac{bc\left(a^2\right)}{bc}=a^2$