Question

Write the value of $\frac{\sin A+\sin 3A}{\cos A+\cos 3A}$.

Answer 1

$$\begin{gathered} \frac{\sin A+\sin 3A}{\cos A+\cos 3A} \\ =\frac{2\sin \left(\frac{A+3A}{2}\right)\cos \left(\frac{A-3A}{2}\right)}{2\cos \left(\frac{A+3A}{2}\right)\cos \left(\frac{A-3A}{2}\right)}\left[\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right),\mathrm{and}\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right] \\ =\frac{\sin 2A\cos \left(-A\right)}{\cos 2A\cos \left(-A\right)} \\ =\frac{\sin 2A\cos A}{\cos 2A\cos A} \\ =\tan 2A \end{gathered}$$