Question

If $\sin \alpha +\sin \beta =\frac{1}{4}$ and $\cos \alpha +\cos \beta =\frac{1}{3}$, then the value of $\cos (\alpha +\beta )$ is,

• A. $\frac{12}{25}$
• B. $\frac{7}{25}$
• C. $\frac{12}{13}$
• D. none of these

Answer 1

The correct option is B $\frac{7}{25}$

$\sin \alpha +\sin \beta =\frac{1}{4}.....\left(1\right)$

$\Rightarrow 2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)=\frac{1}{4}$

$\cos \alpha +\cos \beta =\frac{1}{3}..........\left(2\right)$

$\Rightarrow 2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)=\frac{1}{3}$

Dividing $\left(1\right)$ by $\left(2\right)$, we get

$\frac{2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)}{2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)}=\frac{3}{4}$

$\Rightarrow \tan \left(\frac{\alpha +\beta }{2}\right)=\frac{3}{4}$

Using $\cos 2x=\frac{1-ta{n}^{2}x}{1+{\tan }^{2}x}\cos 2x=\frac{1-ta{n}^{2}x}{1+{\tan }^{2}x}$

$\cos (\alpha +\beta )=\frac{1-{\left(\frac{3}{4}\right)}^2}{1+{\left(\frac{3}{4}\right)}^2}=\frac{7}{25}$