Q59 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics
If $\cos \alpha+\cos \beta=\frac{1}{3}$ and $\sin \alpha+\sin \beta=\frac{1}{4}$, then $\cos (\alpha+\beta)=$
- A$\frac{24}{25}$
- Correct answerB$\frac{7}{25}$
- C$\frac{13}{25}$
- D$\frac{12}{13}$
Solution
Given $\cos \alpha+\cos \beta=\frac{1}{3}$
$$
\Rightarrow 2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)=\frac{1}{3}.....(i)
$$
$\& \sin \alpha+\sin \beta=\frac{1}{4}$
$$
\Rightarrow 2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)=\frac{1}{4}.....(ii)
$$
Eqn. (ii) divided by (i)
$$
\tan \left(\frac{\alpha+\beta}{2}\right)=\frac{3}{4}
$$
Now, $\sin (\alpha+\beta)=\frac{2 \tan \left(\frac{\alpha+\beta}{2}\right)}{1+\tan ^2\left(\frac{\alpha+\beta}{2}\right)}=\frac{2 \times \frac{3}{4}}{1+\left(\frac{3}{4}\right)^2}=\frac{24}{25}$
Hence, the correct answer is option: 'B'.