Question

In any triangle ABC,
$\frac{1-\cos A+\cos B+\cos C}{1-\cos C+\cos A+\cos B}=\frac{\tan \left(A/2\right)}{\tan \left(C/2\right)}$.

• A. True
• B. False

Answer 1

The correct option is A True

$\frac{1-\cos A+\cos B+\cos C}{1-\cos C+\cos A+\cos B}$ $A+B+C=180\Rightarrow A+B=180-C\Rightarrow \cos (A+B)=-\cos C$

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

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

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

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

$=\frac{\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}{\sin \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}}$

$=\frac{\tan \frac{A}{2}}{\tan \frac{C}{2}}$