If \[ \cos A+\cos B=\frac12 \] and \[ \sin A+\sin B=\frac14 \] prove that \[ \tan\frac{A+B}{2}=\frac12 \]
Solution
Given:
\[ \cos A+\cos B=\frac12 \] \[ \sin A+\sin B=\frac14 \]Use identities:
\[ \cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \]
\[
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
=
\frac12
\]
\[
2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
=
\frac14
\]
Divide second equation by first equation:
\[ \frac{ 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} }{ 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} } = \frac{\frac14}{\frac12} \]
\[
\tan\frac{A+B}{2}
=
\frac12
\]
Hence Proved.