Question

$A+B+C={180}^{0}andA=B=Cthentan\frac{A}{2}.\tan \frac{B}{2}+\tan \frac{B}{2}.\tan \frac{C}{2}+tan\frac{A}{2}.tan\frac{C}{2}=$

• A. $1$
• B. $0$
• C. $-1$
• D. None of these

Answer 1

The correct option is A $1$

Given $A+B+C={180}^{\circ }$ and also $A=B=C$

$⟹A=B=C=\frac{{180}^{\circ }}{3}={60}^{\circ }$

$\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}=\left(\tan {30}^{\circ }\right)\left(\tan {30}^{\circ }\right)+\left(\tan {30}^{\circ }\right)\left(\tan {30}^{\circ }\right)+\left(\tan {30}^{\circ }\right)\left(\tan {30}^{\circ }\right)$

$=3{\tan }^2{30}^{\circ }=3\times \frac{1}{3}=1$