Question

Prove that :

$\frac{{\cos }^2\frac{B-C}{2}}{(b+c{)}^2}+\frac{{\sin }^2\frac{B-C}{2}}{(b-c{)}^2}=\frac{1}{a^2}$

Answer 1

we use half angle formulae.

Given,

$\frac{{\cos }^2\frac{B-C}{2}}{(b+c{)}^2}+\frac{{\sin }^2\frac{B-C}{2}}{(b-c{)}^2}$

$\cos \frac{B-C}{2}=\frac{b+c}{a}\sin \frac{A}{2}$

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

substituting these values in given equation, we get,

$=\frac{{\left[\frac{b+c}{a}\sin \frac{A}{2}\right]}^2}{(b+c{)}^2}+\frac{{\left[\frac{b-c}{a}\cos \frac{A}{2}\right]}^2}{(b-c{)}^2}$

$=\frac{1}{a^2}{\sin }^2\frac{A}{2}+\frac{1}{a^2}{\cos }^2\frac{A}{2}$

$=\frac{1}{a^2}({\sin }^2\frac{A}{2}+{\cos }^2\frac{A}{2})$

$=\frac{1}{a^2}$