Question

In a $△ABC,\sum \frac{(1-\tan \frac{B}{4})(1-\tan \frac{C}{4})}{(1+\tan \frac{B}{4})(1+\tan \frac{C}{4})}=$

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $1$

Answer 1

Answer: (D)

$1$

Take $1=\tan \frac{\pi }{4}$ we have
$\sum \frac{(1-\tan \frac{B}{4})(1-\tan \frac{C}{4})}{(1+\tan \frac{B}{4})(1+\tan \frac{C}{4})}$
$=\sum \frac{(\tan \frac{\pi }{4}-\tan \frac{B}{4})(\tan \frac{\pi }{4}-\tan \frac{C}{4})}{(1+\tan \frac{\pi }{4}\tan \frac{B}{4})(1+\tan \frac{\pi }{4}\tan \frac{C}{4})}$
$=\sum \tan (\frac{\pi }{4}-\frac{B}{4})\tan (\frac{\pi }{4}-\frac{C}{4})$ using the formula $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$
Take $\alpha =\frac{\pi -A}{4}$ $\beta =\frac{\pi -B}{4}$ $\gamma =\frac{\pi -C}{4}$
$\Rightarrow \alpha +\beta +\gamma =\frac{\pi -A}{4}+\frac{\pi -B}{4}+\frac{\pi -C}{4}$
$=\frac{3\pi }{4}-\frac{(A+B+C)}{4}$
$=\frac{3\pi }{4}-\frac{\pi }{4}$
$=\frac{2\pi }{4}=\frac{\pi }{2}$
$\sum \tan \beta \tan \gamma =1$
$\therefore S=\sum \tan \beta \tan \gamma =1$