Question

If $\tan (\frac{\alpha }{2})$ and $\tan (\frac{\beta }{2})$ are the roots of the equation $8x^2-26x+15=0$ then $\cos (\alpha +\beta )$ is equal to-

• A. $-\frac{627}{725}$
• B. $\frac{627}{725}$
• C. $-\frac{725}{625}$
• D. $-1$

Answer 1

The correct option is A $-\frac{627}{725}$

Given equation $8x^2-26x+15=0$
$\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}=\frac{26}{8}$
$\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{15}{8}$
$\tan (\frac{\alpha +\beta }{2})=\frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}$
$=\frac{\frac{26}{8}}{1-\frac{15}{8}}=-\frac{26}{7}$
Now, $\cos (\alpha +\beta )=\frac{1-{\tan }^2\frac{(\alpha +\beta )}{2}}{1+{\tan }^2\frac{\alpha +\beta }{2}}$
$=\frac{1-\frac{676}{49}}{1+\frac{676}{49}}$
$=\frac{-627}{725}$