Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

Then $\left(\frac{c+b}{c-b}\right)\times \tan \frac{A}{2}$ is equal to?

• A. $\tan (\frac{A}{2}+B)$
• B. $\cot (\frac{A}{2}+B)$
• C. $\tan (A+\frac{B}{2})$
• D. none of these

Answer 1

The correct option is A $\tan (\frac{A}{2}+B)$

Using sine rule

$\frac{c+b}{c-b}\times \tan \frac{A}{2}=\frac{\sin C+\sin B}{\sin C-\sin B}\tan \frac{A}{2}\frac{c+b}{c-b}\times \tan \frac{A}{2}=\frac{2\sin \frac{C+B}{2}\cos \frac{C-B}{2}}{2\sin \frac{C-B}{2}\cos \frac{C+B}{2}}\tan \frac{A}{2}\frac{c+b}{c-b}\times \tan \frac{A}{2}=\frac{\cos \frac{A}{2}\cos \frac{C-B}{2}}{\sin \frac{C-B}{2}\sin \frac{A}{2}}\tan \frac{A}{2}\frac{c+b}{c-b}\times \tan \frac{A}{2}=\cot \frac{C-B}{2}\frac{c+b}{c-b}\times \tan \frac{A}{2}=\cot \frac{\pi -A-B-B}{2}\frac{c+b}{c-b}\times \tan \frac{A}{2}=\tan (\frac{A}{2}+B)$