Question

The sides a, b, c (taken in that order) of triangle $ABC$ are in A.P. If $\cos \alpha =\frac{a}{b+c},\cos \beta =\frac{b}{a+b},\cos \gamma =\frac{c}{a+b}then{\tan }^2\left(\frac{\alpha }{2}\right)+{\tan }^2\left(\frac{\gamma }{2}\right)$ is equal.

[Note:All symbols used have usual meaning in triangle $ABC$]

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. None of these

Answer 1

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

$a,b,c$ are sides of $△ABC$ in A.P

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

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

$\cos \beta =\frac{b}{2b}=\frac{1}{2}\Rightarrow \beta ={60}^{\circ }$

$\cos \gamma =\frac{c}{a+b}=\frac{b+d}{2b-d}$

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

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

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

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

Again,$\cos \gamma =\frac{b+d}{2b-d}$

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

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

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

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

Now,${\tan }^2\frac{\alpha }{2}+{\tan }^2\frac{\gamma }{2}=\frac{2d+b}{3b}+\frac{b-2d}{3b}$

$=\frac{2d+b+b-2d}{3b}=\frac{2b}{3b}=\frac{2}{3}$

$\therefore {\tan }^2\frac{\gamma }{2}=\frac{2}{3}$