Question

In any $\Delta$ $ABC$, if $\cos \theta =\frac{a}{b+c},\cos \varphi =\frac{b}{a+c},\cos \Psi =\frac{c}{a+b}$, where $\theta ,\varphi$ and $\Psi$ lie between $0$ and $\pi$, prove that ${\tan }^2\frac{\theta }{2}+{\tan }^2\frac{\varphi }{2}+{\tan }^2\frac{\Psi }{2}=1$

Answer 1

Given, $\cos \theta =\frac{a}{b+c}$

$\Rightarrow \frac{1-{\tan }^2\theta /2}{1+{\tan }^2\theta /2}=\frac{a}{b+c}$

$\Rightarrow {\tan }^2\frac{\theta }{2}=\frac{b+c-a}{a+b+c}$ (i)

Similarly ${\tan }^2\frac{\varphi }{2}=\frac{a+c-b}{a+b+c}$ (ii)

and ${\tan }^2\frac{\Psi }{2}=\frac{a+b-c}{a+b+c}$ (iii)

adding (i), (ii) and (iii), we get

${\tan }^2\frac{\theta }{2}+{\tan }^2\frac{\varphi }{2}+{\tan }^2\frac{\Psi }{2}=\frac{a+b+c}{a+b+c}$

$\Rightarrow {\tan }^2\frac{\theta }{2}+{\tan }^2\frac{\varphi }{2}+{\tan }^2\frac{\Psi }{2}=1$