Question

If $\sin x- \sin y = \frac {1}{2}$ and $\cos x- \cos y = 1$, then $ \tan(x + y) $ is equal to

• A. $\frac {3}{8}$
• B. $-\frac{3}{8}$
• C. $\frac {4}{3}$
• D. $-\frac {4}{3}$

Answer 1

Answer: (C)

Given,
$\sin x-\sin y=\frac{1}{2}\,\,\,\,\,\,\,\,...(i)$
and $ \cos x-\cos y=1 \,\,\,\,\,\,\,\,...(ii)$
$\Rightarrow 2 \cos \frac{x+y}{2} \cdot \sin \frac{x-y}{2}=\frac{1}{2} \,\,\,\,\,\,\,\,...(iii)$
and $-2 \sin \frac{x+y}{2} \cdot \sin \frac{x-y}{2}=1\,\,\,\,\,\,\,\,...(iv)$
On dividing Eq. (iv) by Eq. (iii), we get
$-\tan \left(\frac{x+y}{2}\right)=2$
$\Rightarrow \tan \left(\frac{x+y}{2}\right)=-2\,\,\,\,\,\,\,\,...(v)$
Now, $\tan (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1-\tan ^{2}\left(\frac{x+y}{2}\right)}$
$\left(\because \tan 2\,\theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)$
$=\frac{2(-2)}{1-(-2)^{2}}\,\,\,\,\,\,$[using Eq. (v)]
$=\frac{-4}{1-4}=\frac{-4}{-3}=\frac{4}{3}$