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)]
$=\frac{-4}{1-4}=\frac{-4}{-3}=\frac{4}{3}$