Question

If R denotes circumradius then in $\Delta$ABC, $\frac{b^2-c^2}{2aR}$ is equal to

• A. cos (B - C)
• B. 2sin (B - C)
• C. cos B - cos C
• D. None of these

Answer 1

The correct option is B 2sin (B - C)

consider, $t=\frac{b^2-c^2}{2aR}$
$\Rightarrow t=\frac{4R^2{\sin }^{2}B-4R^2{\sin }^{2}C}{4R^2\sin A}$
$\Rightarrow t=\frac{{\sin }^{2}B-{\sin }^{2}C}{\sin (\pi -B-C)}=\frac{\cos 2C-\cos 2B}{\sin (B+C)}$
$\Rightarrow t=\frac{-2\sin (B+C)\sin (C-B)}{\sin (B+C)}=2\sin (B-C)$
Ans: B