Question

If $\tan \frac{\alpha }{2}$ and $\tan \frac{\beta }{2}$ are the roots of the equation $9x^2-26x+15=0$ then $\cos (\alpha +\beta )$ is equal to?

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

Answer 1

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

$\Rightarrow$ The given equation is $9x^2-26x+15=0$ comparing it with $a{x}^2+bx+c=0$

$\Rightarrow$ We get, $a=9,b=-26,c=15$

$\Rightarrow$ $\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}=\frac{-b}{a}=-\frac{-26}{8}=\frac{13}{4}$ ----- ( 1 )

$\Rightarrow$ $\tan \left(\frac{\alpha }{2}\right)\tan \left(\frac{\beta }{2}\right)=\frac{c}{a}=\frac{15}{8}$ ------- ( 2 )

Now,

$\Rightarrow$ $\tan \left(\frac{\alpha +\beta }{2}\right)=\tan (\frac{\alpha }{2}+\frac{\beta }{2})$

$=\frac{\tan \left(\frac{\alpha }{2}\right)+\tan \left(\frac{\beta }{2}\right)}{1-\tan \left(\frac{\alpha }{2}\right)\tan \left(\frac{\beta }{2}\right)}$

$=\frac{\frac{13}{4}}{1-\frac{15}{8}}$ [ From ( 1 ) and ( 2 ) ]

$=\frac{-26}{7}$

$\therefore$ $\tan \left(\frac{\alpha +\beta }{2}\right)=\frac{-26}{7}$ ---- ( 3 )

$\Rightarrow$ $\cos (\alpha +\beta )=\frac{1-{\tan }^2\left(\frac{\alpha +\beta }{2}\right)}{1+{\tan }^2\left(\frac{\alpha +\beta }{2}\right)}$

$=\frac{1-{\left(\frac{-26}{7}\right)}^2}{1+{\left(\frac{26}{7}\right)}^2}$ [ From ( 3 ) ]

$=\frac{1-\frac{676}{49}}{1+\frac{676}{49}}$

$=\frac{-627}{725}$

$\Rightarrow$ $\cos (\alpha +\beta )=-\frac{627}{725}$