Question

The sides a,b,c(taken in order) of triangle ABC are in A.P. If $\cos \alpha =\frac{a}{b+c},\cos \beta =\frac{b}{c+a},\cos \gamma =\frac{c}{a+b}then{\tan }^2\frac{\alpha }{2}+{\tan }^2\frac{\gamma }{2}$ is equal to [Note: All symbols used have usual meaning in triangle ABC]

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

a,b,c are sides of $△$ ABC in A.P.

$\therefore a=b-d,c=b+d$ is the common difference.

$\cos \alpha =\frac{a}{b+c}=\frac{b-d}{2b+d},\cos \beta =\frac{b}{2b}=\frac{1}{2}$

$\cos \gamma =\frac{c}{a+b}=\frac{b+d}{2b-d},\therefore \beta ={60}^{\circ }$

$\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}=\frac{b-d}{2b+d}$

${\tan }^2\frac{\alpha }{2}=\frac{(2b+d)-(b-d)}{(2b+d)+(b-d)}$

${\tan }^2\frac{\alpha }{2}=\frac{b+2d}{3b}$

$\frac{1-{\tan }^2\frac{\gamma }{2}}{1+{\tan }^2\frac{\gamma }{2}}=\frac{(2b-d)-(b+d)}{(2b-d)+(b+d)}$

${\tan }^2\frac{\gamma }{2}=\frac{b-2d}{3b}$

${\tan }^2\frac{\alpha }{2}+{\tan }^2\frac{\gamma }{2}=\frac{2b}{3b}=\frac{2}{3}$