Question

In triangle $ABC,\frac{\sin A+\sin B+\sin C}{\sin A+\sin B-\sin C}$ is equal to

• A. $\tan \frac{A}{2}\cot \frac{B}{2}$
• B. $\cot \frac{A}{2}\tan \frac{B}{2}$
• C. $\cot \frac{A}{2}\cot \frac{B}{2}$
• D. $\tan \frac{A}{2}\tan \frac{B}{2}$

Answer 1

The correct option is C $\cot \frac{A}{2}\cot \frac{B}{2}$

We know that
$A+B+C=\pi$
Now,
$\sin A+\sin B-\sin C=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}-2\sin \frac{C}{2}\cos \frac{C}{2}(\because \sin \left(\frac{A+B}{2}\right)=\sin (\frac{\pi }{2}-\frac{C}{2})=\cos \frac{C}{2})=2\cos \frac{C}{2}[\cos \frac{A-B}{2}-\sin \frac{C}{2}]=2\cos \frac{C}{2}[\cos \frac{A-B}{2}-\cos \frac{A+B}{2}](\because \sin \left(\frac{C}{2}\right)=\sin (\frac{\pi }{2}-\frac{A+B}{2})=\cos \frac{A+B}{2})=4\sin \frac{A}{2}\sin \frac{B}{2}\cos \frac{C}{2}$

Similarly
$\sin A+\sin B+\sin C=2\cos \frac{C}{2}[\cos \frac{A-B}{2}+\cos \frac{A+B}{2}]=4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

$\therefore \frac{\sin A+\sin B+\sin C}{\sin A+\sin B-\sin C}=\cot \frac{A}{2}\cot \frac{B}{2}$