Question

If $R$ denotes the circum radius of a $\Delta ABC$, then $\frac{b^2-c^2}{2aR}$ is equal to?

• A. $\cos (B-C)$
• B. $\cos B-\cos C$
• C. $\sin (B-C)$
• D. $\sin B-\sin C$

Answer 1

The correct option is C $\sin (B-C)$

we can solve for equilateral traingle

$a=2R\tan A$

$b=2R\tan B$

$c=2R\tan C$

now put these in first equaation

$\frac{4R^2(\tan B-\tan C)}{4R^2\tan A}$

$\frac{\sin (B-C)}{\tan A\cos B\cos C}$

now we know that denminator turns to 1

Hence ans is $\sin (B-C)$