Question

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. $\frac{-8}{3}$
• D. $\frac{4}{3}$

Answer 1

Answer: (D)

$\frac{4}{3}$

It is given that:

$\sin x+\sin y=\frac{1}{2}.........\left(1\right)\cos x+\cos y=1.........\left(2\right)$

Equation 1 implies:

$2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=\frac{1}{2}.........\left(3\right)$ and

Equation 2 implies:

$2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=1.........\left(4\right)$

Dividing equation 3 and 4, we get, $\tan \left(\frac{x+y}{2}\right)=\frac{1}{2}$

Now consider,

$\tan (x+y)=\frac{2\tan \left(\frac{x+y}{2}\right)}{1-{\tan }^2\left(\frac{x+y}{2}\right)}=\frac{2\left(\frac{1}{2}\right)}{1-{\left(\frac{1}{2}\right)}^2}=\frac{1}{\frac{4-1}{4}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$

Hence, $\tan (x+y)=\frac{4}{3}$.