Question

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

• A. $\frac{627}{725}$
• B. $-\frac{627}{725}$
• C. $-1$
• D. None of these

Answer 1

Answer: (B)

The given equation is $8x^2-26x+15=0$.
$\therefore$ The sum of roots,
$\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}=\frac{26}{8}=\frac{13}{4}$ and product of roots,
$\tan \frac{\alpha }{2}\cdot \tan \frac{\beta }{2}=\frac{15}{8}$
$\therefore \tan \left(\frac{\alpha +\beta }{2}\right)=\frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\cdot \tan \frac{\beta }{2}}$
$=\frac{\frac{13}{4}}{1-\frac{15}{8}}=-\frac{26}{7}$
Now, $\cos (\alpha +\beta )=\frac{1-{\tan }^2\left(\frac{\alpha +\beta }{2}\right)}{1+{\tan }^2\left(\frac{\alpha +\beta }{2}\right)}$
$\left[\because \cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right]$
$=\frac{1-{\left(-\frac{26}{7}\right)}^2}{1+{\left(-\frac{26}{7}\right)}^2}=\frac{49-676}{49+676}$
$=-\frac{627}{725}$