Question

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

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

Answer 1

The correct option is A $0$

$\sum \frac{(1-\tan B/u)(1-\tan \frac{c}{u})}{(1+\tan B/u)(1+\tan \frac{c}{u})}$

Put $C=\pi ,B=\pi$

$\frac{(1-\tan \pi /4)(1-\tan \pi /4)}{(1+\tan \pi /4)(1+\tan \pi /4)}$

$=\frac{(1-1)(1-1)}{(1+2)(1+1)}=\frac{0\times 0}{2\times 2}=0$