Question

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

• A. $\frac{627}{725}$
• B. $\frac{547}{715}$
• C. $-\frac{547}{715}$
• D. $-\frac{627}{725}$

Answer 1

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

$\tan \frac{\alpha }{2},\tan \frac{\beta }{2}$ are the roots of $8x^2-26x+15=0$
Sum and product of roots,
$\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}=\frac{26}{8}=\frac{13}{4}\tan \frac{\alpha }{2}\cdot \tan \frac{\beta }{2}=\frac{15}{8}\tan (\frac{\alpha }{2}+\frac{\beta }{2})=\frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\cdot \tan \frac{\beta }{2}}=\frac{\frac{13}{4}}{1-\frac{15}{8}}=-\frac{26}{7}\because \cos \theta =\frac{1-{\tan }^2\theta /2}{1+{\tan }^2\theta /2}\cos (\alpha +\beta )=\frac{1-{\tan }^2(\frac{\alpha }{2}+\frac{\beta }{2})}{1+{\tan }^2(\frac{\alpha }{2}+\frac{\beta }{2})}=\frac{1-{\left(\frac{-26}{7}\right)}^2}{1+{\left(\frac{-26}{7}\right)}^2}=-\frac{627}{725}$