# Question #d4c2f

Aug 28, 2016

We know from properties of triangle that for $\Delta A B C$

$\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C = 2 R \ldots \left(1\right)$

where a,b,c are three sides of $\Delta A B C$ opposite to angles A,B,C respectively and R is the circum radius of $\Delta A B C$

By relation (1)

$\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C = 2 R$

$\implies a = 2 R \sin A , b = 2 R \sin B , c = 2 R \sin C$

Now

$R H S = \frac{{a}^{2} - {b}^{2}}{c} ^ 2 = \frac{4 {R}^{2} \left({\sin}^{2} A - {\sin}^{2} B\right)}{4 {R}^{2} {\sin}^{2} C}$

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

$= \frac{\sin \left(A + B\right) \sin \left(A - B\right)}{\sin} ^ 2 \left(A + B\right)$
^
$= \sin \frac{A - B}{\sin} \left(A + B\right) = L H S$

Proved