Question

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

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

Answer 1

The correct option is B

$-\frac{627}{725}$

Explanation for the correct option:

Step 1: Find the value of the sum and product of the roots of the given equation.

Given : $8x^2-26x+15=0$

Comparing the given equation with $\mathit{a}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{b}\mathit{x}\mathbf{+}\mathit{c}\mathbf{=}\mathbf{0}$

The sum of the roots $=-\frac{b}{a}$ and The product of the roots $=\frac{c}{a}$

The sum of roots: $\tan \left(\frac{\alpha }{2}\right)+\tan \left(\frac{\beta }{2}\right)=\frac{26}{8}$

$\mathbf{Or},\tan \left(\frac{\alpha }{2}\right)+\tan \left(\frac{\beta }{2}\right)=\frac{13}{4}$

And the product of roots:$\tan \left(\frac{\alpha }{2}\right)\tan \left(\frac{\beta }{2}\right)=\frac{15}{8}$

Step 2: Calculate the value of $\tan \left(\frac{\alpha +\beta }{2}\right)$

$\begin{array}{rcl}\therefore \tan \left(\frac{\alpha +\beta }{2}\right)& =& \frac{\tan \left({\displaystyle \frac{\alpha }{2}}\right)+\tan \left({\displaystyle \frac{\beta }{2}}\right)}{1-\tan \left({\displaystyle \frac{\alpha }{2}}\right)\tan \left({\displaystyle \frac{\beta }{2}}\right)}\\ & =& \frac{{\displaystyle \frac{13}{4}}}{1-{\displaystyle \frac{15}{8}}}\\ & =& -\frac{26}{7}\end{array}$

Step 3: Finding the value of $\cos (\alpha +\beta )$

$\begin{array}{rcl}\cos \left(\alpha +\beta \right)& =& \frac{1-{\tan }^2\left({\displaystyle \frac{\alpha +\beta }{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{\alpha +\beta }{2}}\right)}\left[\because \cos \left(2\theta \right)=\frac{1-{\tan }^2\left(\theta \right)}{1+{\tan }^2\left(\theta \right)}\right]\\ & =& \frac{1-{\left(-{\displaystyle \frac{26}{7}}\right)}^2}{1+{\left(-{\displaystyle \frac{26}{7}}\right)}^2}\\ & =& \frac{49-676}{49+676}\\ & =& -\frac{627}{725}\end{array}$

Hence, the correct option is (B).