If A + B + C = π, prove that

Solution:

$=\cos 2 A-\cos 2 B+\cos 2 C$

Using,

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

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

$=\cos 2 A-\{2 \sin (B+C) \sin (B-C)\}$

since $A+B+C=\pi$

$\rightarrow \mathrm{B}+\mathrm{C}=180-\mathrm{A}$

And $\sin (\pi-A)=\sin A$

$=\cos 2 A-\{2 \sin (\pi-A) \sin (B-C)\}$

$=\cos 2 A-\{2 \sin A \sin (B-C)\}$

$=\cos 2 A-2 \sin A \sin (B-C)$

Using, $\cos 2 A=1-2 \sin ^{2} A$

$=-2 \sin ^{2} A+1-2 \sin A \sin (B-C)$

$=-2 \sin A\{\sin A+\sin (B-C)\}+1$

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

$=-2 \sin A\left\{2 \sin \left(\frac{A+B-C}{2}\right) \cos \left(\frac{A+C-B}{2}\right)\right\}+1$

$=-2 \sin A\left\{2 \sin \left(\frac{\pi-C-C}{2}\right) \cos \left(\frac{\pi-B-B}{2}\right)\right\}+1$

$=-2 \sin A\left\{2 \sin \left(\frac{\pi}{2}-\frac{2 C}{2}\right) \cos \left(\frac{\pi}{2}-\frac{2 B}{2}\right)\right\}+1$

As, $\sin \left(\frac{\pi}{2}-\mathrm{A}\right)=\cos \mathrm{A}$

$=-2 \sin A\{2 \cos C \sin B\}+1$

$=-4 \sin A \cos B \sin C+1$

= R.H.S