a, b, c are the sides of a scalene triangle A | vedclass

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(C) Given $\cos \alpha = \frac{a}{b+c}$.Using the identity $\cos \alpha = \frac{1 - 	an^2(\alpha/2)}{1 + 	an^2(\alpha/2)}$,we have $\frac{1 - 	an^2

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(C) Given $\cos \alpha = \frac{a}{b+c}$.Using the identity $\cos \alpha = \frac{1 - 	an^2(\alpha/2)}{1 + 	an^2(\alpha/2)}$,we have $\frac{1 - 	an^2(\alpha/2)}{1 + 	an^2(\alpha/2)} = \frac{a}{b+c}$.Applying componendo and dividendo,$\frac{(1 + 	an^2(\alpha/2)) + (1 - 	an^2(\alpha/2))}{(1 + 	an^2(\alpha/2)) - (1 - 	an^2(\alpha/2))} = \frac{(b+c) + a}{(b+c) - a}$.This simplifies to $\frac{2}{2 	an^2(\alpha/2)} = \frac{a+b+c}{b+c-a}$,so $	an^2(\alpha/2) = \frac{b+c-a}{a+b+c}$.Similarly,$	an^2(\beta/2) = \frac{c+a-b}{a+b+c}$ and $	an^2(\gamma/2) = \frac{a+b-c}{a+b+c}$.Summing these,$	an^2(\alpha/2) + 	an^2(\beta/2) + 	an^2(\gamma/2) = \frac{(b+c-a) + (c+a-b) + (a+b-c)}{a+b+c} = \frac{a+b+c}{a+b+c} = 1$.

$a, b, c$ are the sides of a scalene triangle $ABC$. 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 $\tan^2 \frac{\alpha}{2} + \tan^2 \frac{\beta}{2} + \tan^2 \frac{\gamma}{2} =$

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