Question

In a $\Delta ABC,\Sigma (b+c)\tan \frac{A}{2}\tan \left(\frac{B-C}{2}\right)$ is equal to

• A. a
• B. b
• C. c
• D. 0

Answer 1

Answer: (D)

$\Sigma (b+c)\tan \frac{A}{2}\tan \left(\frac{B-C}{2}\right)$
$\therefore (b+c)\tan \frac{A}{2}\tan \left(\frac{B-C}{2}\right)$
$=(b+c)\cdot \frac{(b-c)}{(b+c)}\cot \frac{A}{2}\tan \frac{A}{2}$
$=b-c$
$\therefore \Sigma (b+c)\tan \frac{B-C}{2}\tan \frac{A}{2}$
$=b-c+c-a+a-b=0$