Find the values of $\cos(\alpha+\beta) $ if the roots of an equation are given in terms of tan It is given that $ \tan\frac{\alpha}{2} $ and $ \tan\frac{\beta}{2} $ are the zeroes of the equation $ 8x^2-26x+15=0$ then find the value of $\cos(\alpha+\beta$).
I attempted to solve this but I don't know if my solution is right.Can someone verify this.
sum of roots = $ \frac{-b}{a}$
so  $ \tan\frac{\alpha}{2} + \tan\frac{\beta}{2} $ = $ \frac{26}{8}$
product of roots = $ \frac{c}{a}$
so  $ \tan\frac{\alpha}{2} $ . $ \tan\frac{\beta}{2} $ = $ \frac{15}{8}$
$$ \tan(\frac{\alpha+\beta}{2} ) = \frac{\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}}{1-\tan\frac{\alpha}{2}.\tan\frac{\beta}{2}}$$
Puting the values in :-
$$ \tan(\frac{\alpha+\beta}{2} ) = -\frac{26}{7}$$
$$ \sec^2(\frac{\alpha+\beta}{2}) = \tan^2(\frac{\alpha+\beta}{2} ) + 1 $$
$$ \sec^2(\frac{\alpha+\beta}{2}) = \frac{(-26)^2 + 7^2}{7^2} $$
$$ \cos^2{\frac{\alpha+\beta}{2}} = \frac{7^2}{25^2}$$
$$ \frac{\cos(\alpha+\beta) + 1}{2} = \frac{49}{725}$$
$$ \cos(\alpha+\beta) = \frac{-627}{725}$$
 A: Since, $\tan\frac{\alpha}{2}$ & $\tan\frac{\beta}{2}$ are roots of the equation $8x^2-26x+15=0$ Hence, we have $$\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}=\frac{-(-26)}{8}=\frac{13}{4}$$
$$\tan\frac{\alpha}{2}\tan\frac{\beta}{2}=\frac{15}{8}$$
$$\implies \tan \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\frac{\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}}{1-\tan\frac{\alpha}{2}\tan\frac{\beta}{2}}$$
$$\implies \tan \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\frac{\frac{13}{4}}{1-\frac{15}{8}}$$  $$\implies \color{red}{\tan \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=-\frac{26}{7}}$$
Now, we have
$$\cos (\alpha+\beta)=\cos2\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)$$
$$=\frac{1-\tan^2\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)}{1+\tan^2\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)}$$
$$=\frac{1-\left(\frac{-26}{7}\right)^2}{1+\left(\frac{-26}{7}\right)^2}$$ $$\implies \color{blue}{\cos(\alpha+\beta)=-\frac{627}{725}}$$
The answer above is same as you obtained. Your answer is correct.
A: Yes, your answer and method, all are right. Just $\tan(\frac{a+b}{2})=\frac{\tan(\frac a2)+\tan(\frac b2)}{1-\tan(\frac a2)\cdot\tan(\frac b2)}$. 
But your values are right. And this was the slight correction.
