Question

In a $\Delta ABC,\frac{bc{\sin }^{2}A}{\cos A+\cos B\cos C}$ is equal to

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

Answer 1

The correct option is C $a^2$

In $△ABC$

$\frac{bc{\sin }^{2}A}{\cos A+\cos B\cdot \cos C}$

$=\frac{2bc{\sin }^{2}A}{2\cos A+2\cos B\cdot \cos C}$

$=\frac{2bc{\sin }^{2}A}{2\cos A+\cos (B+C)+\cos (B-C)}[\because 2\cos x\cos y=\cos (x+y)+\cos (x-y)]$

$=\frac{2bc{\sin }^{2}A}{2\cos A+\cos (\pi -A)+\cos (B-C)}[\because A+B+C=\pi ]$

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

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

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

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

$=\frac{2bc{\sin }^{2}A}{2\sin B\sin C}[\because \cos x-\cos y=-2\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)]$

$=\frac{a^2{\sin }^{2}A}{{\sin }^{2}A}\left[\text{using}\sin \text{law}\right]$

$=a^2$