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Q.
If $\sin\,x+\sin\,y=\frac{7}{5}$ and $\cos\,x+\cos\,y=\frac{1}{5}$, then $\sin(x+y)$ equals
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Solution:
We have,
$\sin\,x+\sin\,y=\frac{7}{5}$
and $\cos\,x+\cos\,y=\frac{1}{5}$
$\Rightarrow 2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=\frac{7}{5} \ldots\left(i\right)$
and $2\cos\left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=\frac{1}{5} \ldots\left(ii\right)$
On dividing Eq. $\left(ii\right)$ by Eq. $\left(i\right)$, we get
$\tan \left(\frac{x+y}{2}\right)=7 $
$\Rightarrow \sin\left(x+y\right)=\frac{2\,\tan\left(\frac{x+y}{2}\right)}{1+\tan^{2}\left(\frac{x+y}{2}\right)}$
$\Rightarrow \sin \left(x+y\right)=\frac{14}{1+49}=\frac{14}{50}$
$\Rightarrow \sin\left(x+y\right)=\frac{7}{25}$