Question

If in a $\Delta ABC,\frac{r}{r_1}=\frac{1}{2},$ then the value of $tan\frac{A}{2}(tan\frac{B}{2}+tan\frac{C}{2})$ is equal to :

• A. 2
• B. $\frac{1}{2}$
• C. 1
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{2}$

$\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}=1......\left(1\right)$

Given,

$\frac{r}{r_1}=\frac{1}{2}$

$\frac{4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}{4R\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}=\frac{1}{2}$

$\tan \frac{B}{2}\tan \frac{C}{2}=\frac{1}{2}.......\left(2\right)$

$\left(2\right)$ in $\left(1\right)$

$\tan \frac{A}{2}\tan \frac{B}{2}+\frac{1}{2}+\tan \frac{C}{2}\tan \frac{A}{2}=1$

$\tan \frac{A}{2}(\tan \frac{B}{2}+\tan \frac{C}{2})=1-\frac{1}{2}$

$=\frac{1}{2}$