Solve this


If $\cos x+\cos y=\frac{1}{3}$ and $\sin x+\sin y=\frac{1}{4}$, prove that $\tan \left(\frac{x+y}{2}\right)=\frac{3}{4}$



$\cos x+\cos y=\frac{1}{3}$ .............$-i$

$\sin x+\sin y=\frac{1}{4}$ ................-ii

dividing ii by I we get,

$\Rightarrow \frac{\sin x+\sin y}{\cos x+\cos y}=\frac{\frac{1}{4}}{\frac{1}{3}}$

$\Rightarrow \frac{\sin x+\sin y}{\cos x+\cos y}=\frac{3}{4}$

$\Rightarrow \frac{2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}}{2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}}=\frac{3}{4}$

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

Using the formula,

$\sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$

$\cos A+\cos B=2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now