Question

In $\Delta ABC$, if 2b=a+c show that 3 tan $\left(\frac{A}{2}\right)tan\left(\frac{C}{2}\right)=1$.

Answer 1

In $\Delta$ABC

$2b=a+c$ $[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R]$

$2\times \left(2R\sin B\right)=2R(\sin A+\sin C)$

$2\sin B=\sin A+\sin C$ $[\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}]$

$2\times 2\sin \frac{B}{2}\cos \frac{B}{2}=2\sin \frac{A+C}{2}\cos \frac{A-C}{2}$

$\Rightarrow 2\sin \frac{B}{2}\cos \frac{B}{2}=\sin \left(\frac{\pi -\beta }{2}\right)\cos \frac{A-C}{2}$

$\Rightarrow 2\sin \frac{\pi -(A+C)}{2}\cos \frac{B}{2}=\cos \frac{B}{2}\cos \frac{A-C}{2}$

$\Rightarrow 2\cos \frac{A+C}{2}=\cos \frac{A-C}{2}$

$\Rightarrow \frac{2}{1}=\frac{\cos \frac{A-C}{2}}{\cos \frac{A+C}{2}}$

Now applying component dividend

$\frac{2+1}{2-1}=\frac{\cos \frac{A-C}{2}+\cos \frac{A+C}{2}}{\cos \frac{A-C}{2}-\cos \frac{A+C}{2}}$ $[\frac{a}{b}=\frac{c}{d},then\frac{a+b}{a-b}=\frac{c+d}{c-d}]$

$\frac{3}{1}=\frac{2\cos \left(\frac{A}{2}\right)\cos \left(\frac{C}{2}\right)}{2\sin \frac{A}{2}\sin \frac{C}{2}}$ $[\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}]$

$3=\frac{1}{\tan \frac{A}{2}\tan \frac{C}{2}}$ $[\cos \alpha -\cos \beta =2\sin \frac{\alpha +\theta }{2}\sin \frac{\beta -\alpha }{2}]$

$3\tan \frac{A}{2}\tan \frac{C}{2}=1$.