\(a, b, c\) are the sides of a scalene triangle \(A B C\). If angles \(\alpha, \beta, \gamma\) lie between 0…
\(a, b, c\) are the sides of a scalene triangle \(A B C\). If angles \(\alpha, \beta, \gamma\) lie between 0 and \(\pi\) such that \(\cos \alpha=\frac{a}{b+c}, \cos \beta=\frac{b}{c+a}\) and \(\cos \gamma=\frac{c}{a+b}\), then \(\tan ^2 \frac{\alpha}{2}+\tan ^2 \frac{\beta}{2}+\tan ^2 \frac{\gamma}{2}=\)
\(\frac{1}{3}\)
2
1
\(\frac{3}{2}\)
Solution
It is given \(\cos \alpha=\frac{a}{b+c}\)
\(\Rightarrow \quad \frac{1-\tan ^2 \frac{\alpha}{2}}{1+\tan ^2 \frac{\alpha}{2}}=\frac{a}{b+c}\)
On applying componendo and dividendo law, we get
\(\frac{2 \tan ^2 \alpha / 2}{2}=\frac{b+c-a}{b+c+a}\)
\(\Rightarrow \quad \tan ^2 \alpha / 2=\frac{b+c-a}{a+b+c}\)
Similarly, \(\quad \tan ^2 \frac{\beta}{2}=\frac{c+a-b}{a+b+c}\)
and \(\tan ^2 \frac{\gamma}{2}=\frac{a+b-c}{a+b+c}\)
So, \(\tan ^2 \frac{\alpha}{2}+\tan ^2 \frac{\beta}{2}+\tan ^2 \frac{\gamma}{2}=\frac{a+b+c}{a+b+c}=1\)
Hence, option (c) is correct.