Question

In triangle $ABC$; $1-\left(\tan \frac{B}{2}\tan \frac{C}{2}\right)$ is equal to -

• A. $\frac{2a}{a+b+c}$
• B. $\frac{2b}{a+b+c}$
• C. $\frac{2c}{a+b+c}$
• D. $noneofthese$

Answer 1

The correct option is A $\frac{2a}{a+b+c}$

We have,

$1-\tan \frac{B}{2}\tan \frac{C}{2}$

$=1-\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$

$=1-\sqrt{\frac{(s-a)(s-c)(s-a)(s-b)}{s(s-b)s(s-c)}}$

$=1-\frac{s-a}{s}$

$=1-\frac{\frac{a+b+c}{2}-a}{\frac{a+b+c}{2}}$

$=1-\frac{\frac{a+b+c-2a}{2}}{\frac{a+b+c}{2}}$

$=1-\frac{a+b+c-2a}{a+b+c}$

$=\frac{a+b+c-a-b-c+2a}{a+b+c}$

$=\frac{2a}{a+b+c}$

Hence, this is the answer.