Question

If $A,B$ and $C$ are the angles of a triangle, then $\tan (\frac{B}{4}+\frac{C}{4})$ equals to

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

Answer 1

The correct option is A $\frac{\cos \left(\frac{A}{2}\right)}{1+\sin \left(\frac{A}{2}\right)}$

$\tan (\frac{B}{4}+\frac{C}{4})$

as $A,B,C$ are angles of triangle,$A+B+C=\pi$

hence, $\frac{A+B+C}{4}=\frac{\pi }{4}$

$\frac{B}{4}+\frac{C}{4}=\frac{\pi }{4}-\frac{A}{4}$

$=\frac{\tan \frac{\pi }{4}-\tan \frac{A}{4}}{1+\tan \frac{\pi }{4}\tan \frac{A}{4}}$
$=\frac{1-\tan \frac{A}{4}}{1+\tan \frac{A}{4}}=\frac{\cos \frac{A}{4}-\sin \frac{A}{4}}{\cos \frac{A}{4}+\sin \frac{A}{4}}$
$=\frac{{\cos }^2\frac{A}{4}-{\sin }^2\frac{A}{4}}{1+2\sin \frac{A}{4}\cos \frac{A}{4}}=\frac{\cos \frac{A}{2}}{1+\sin \frac{A}{2}}$