Question

If $A+B+C=\pi$ & $\sin (A+\frac{C}{2})=k\sin \frac{C}{2}$, then $\tan \frac{A}{2}\tan \frac{B}{2}=$

• A. $\frac{k-1}{k+1}$
• B. $\frac{k+1}{k-1}$
• C. $\frac{k}{k+1}$
• D. $\frac{k+1}{k}$

Answer 1

Answer: (A)

$\frac{k-1}{k+1}$

$\frac{C}{2}=\frac{x-(A+B)}{2}=\frac{\pi }{2}-\frac{(A+B)}{2}$

$\sin (A+\frac{C}{2})=k\sin \frac{C}{2}$

$\sin (A+\frac{\pi }{2}-\frac{A}{2}-\frac{B}{2})=k\sin (\frac{\pi }{2}-\frac{A}{2}-\frac{B}{2})$

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

after expansion

$\frac{\cos \frac{A}{2}.\cos \frac{B}{2}+\sin \frac{A}{2}.\sin \frac{B}{2}}{\cos \frac{A}{2}.\cos \frac{B}{2}-\sin \frac{A}{2}.\sin \frac{B}{2}}=K$

$\frac{\frac{\cos \frac{A}{2}.\cos \frac{B}{2}+\sin \frac{A}{2}.\sin \frac{B}{2}}{\cos \frac{A}{2}.\cos \frac{B}{2}}}{\frac{\cos \frac{A}{2}.\cos \frac{B}{2}-\sin \frac{A}{2}.\sin \frac{B}{2}}{\cos \frac{A}{2}.\cos \frac{B}{2}}}=\frac{k+1}{k-1}$

$\frac{1+\tan \frac{A}{2}.\tan \frac{B}{2}}{1-\frac{A}{2}.\tan \frac{B}{2}}=k\Rightarrow \tan \frac{A}{2}.\tan \frac{A}{2}=\frac{k-1}{k+1}$