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

If $\cos \alpha+\cos \beta=\frac{1}{3}$ and $\sin \alpha+\sin \beta=\frac{1}{4}$, then $\cos (\alpha+\beta)=$
  1. $\frac{24}{25}$
  2. $\frac{7}{25}$
  3. $\frac{13}{25}$
  4. $\frac{12}{13}$

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, $\sin (\alpha+\beta)=\frac{2 \tan \left(\frac{\alpha+\beta}{2}\right)}{1+\tan ^2\left(\frac{\alpha+\beta}{2}\right)}=\frac{2 \times \frac{3}{4}}{1+\left(\frac{3}{4}\right)^2}=\frac{24}{25}$

Asked in: AP EAMCET 2023 (18 May Shift 2)

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