Q115 / 180 · KCET 2015 · Mathematics
If \( \sin x+\sin y=\frac{1}{2} \) and \( \cos x+\cos y=1 \), then \( \tan (x+y)= \)
- A\( \frac{8}{3} \)
- B\( -\frac{3}{4} \)
- C\( -\frac{8}{3} \)
- Correct answerD\( \frac{4}{3} \)
Solution
Given that, $\sin x+\sin y=\frac{1}{2} \rightarrow(1)$
$\cos x+\cos y=1 \rightarrow(2)$
From Eqs. (1) and (2), we have
$2 \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2}=\frac{1}{2} \rightarrow(3)$ $2 \cos \frac{x+y}{2} \cdot \cos \frac{x-y}{2}=1 \rightarrow(4)$
Dividing Eq. (3) by Eq. (4), we have
$\tan \frac{x+y}{2}=\frac{1}{2}$
We know that $\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}$
So, $\tan (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1-\tan ^{2}\left(\frac{x+y}{2}\right)}$
$\Rightarrow \tan (x+y)=\frac{2\left(\frac{1}{2}\right)}{1-\left(\frac{1}{2}\right)^{2}}=\frac{4}{3}$
Hence, the correct answer is option: 'D'.