Question

If $\tan \frac{\alpha}{2}$, and $\tan \frac{\beta}{2}$ are the roots of $8 x^{2}-26 x+15=0$, 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 $8 x^{2}-26 x+15=0$.
$\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, $\cos (\alpha+\beta) =\frac{1-\tan ^{2}\left(\frac{\alpha+\beta}{2}\right)}{1+\tan ^{2}\left(\frac{\alpha+\beta}{2}\right)}$
$\left[\because \cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right]$
$=\frac{1-\left(-\frac{26}{7}\right)^{2}}{1+\left(-\frac{26}{7}\right)^{2}}=\frac{49-676}{49+676}$
$=-\frac{627}{725}$