Question

In a $\Delta ABC$,
$\frac{bsin(C-A)}{c^2-a^2}+\frac{csin(A-B)}{a^2-b^2}=$

• A. $\frac{1}{2R}$
• B. $\frac{1}{R}$
• C. $\frac{2}{R}$
• D. none of these

Answer 1

The correct option is B $\frac{1}{R}$

let $t=\frac{bsin(C-A)}{c^2-a^2}+\frac{csin(A-B)}{a^2-b^2}$

$\Rightarrow t=\frac{2R\sin Bsin(C-A)}{4R^2({\sin }^{2}C-{\sin }^{2}A)}+\frac{2R\sin Csin(A-B)}{4R^2({\sin }^{2}A-{\sin }^{2}B)}$

$\Rightarrow t=\frac{1}{R}[\frac{\sin (\pi -A-C)sin(C-A)}{\cos 2A-\cos 2C}+\frac{\sin (\pi -A-B)sin(A-B)}{\cos 2B-\cos 2A}]$

$\Rightarrow t=\frac{1}{R}[\frac{\sin (A+C)sin(C-A)}{2\sin (C+A)\sin (C-A)}+\frac{\sin (A+B)sin(A-B)}{2\sin (A+B)\sin (A-B)}]$

$\Rightarrow t=\frac{1}{R}$
Ans: B