Question

TanA/2*TanB/2+TanB/2*TanC/2+TanC/2*TanA/2=1

Answer 1

I think this will help you.

Answer 2

Step-by-step explanation:

Let us consider the angles in the triangle ABC, where we know that sum of angles in a triangle is 180°

$A+B+C=180^{\circ}=\pi$

Dividing both sides with 2

$\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}$

$\frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}$

Consider tan trigonometric function on both sides.

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

As Π/2 - C/2 will be in first quadrant and tan (Π/2 – x) = cotx

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

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

We know that,

$\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \cdot \tan y}$

On substituting, we get,

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

On cross multiplying, we get,

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

Therefore,

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

Hence, proved.