Question

Prove that:

$\frac{\sin \mathrm{\theta }+\sin 2\mathrm{\theta }}{1+\cos \mathrm{\theta }+\cos 2\mathrm{\theta }}=\tan \mathrm{\theta }$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{\mathrm{sin\theta }+\sin 2\mathrm{\theta }}{1+\mathrm{cos\theta }+\cos 2\mathrm{\theta }} \\ =\frac{\mathrm{sin\theta }+\sin 2\mathrm{\theta }}{\mathrm{cos\theta }+\left(1+\cos 2\mathrm{\theta }\right)} \end{gathered}$$

$=\frac{\mathrm{sin\theta }+2\mathrm{sin\theta }\mathrm{cos\theta }}{\mathrm{cos\theta }+2{\cos }^2\mathrm{\theta }}\left[\because \sin 2\mathrm{\theta }=2\mathrm{sin\theta cos\theta }\mathrm{and}2{\cos }^2\mathrm{\theta }=1+\cos 2\mathrm{\theta }\right]$

$$\begin{gathered} =\frac{\mathrm{sin\theta }\left(1+2\mathrm{cos\theta }\right)}{\mathrm{cos\theta }\left(1+2\mathrm{cos\theta }\right)} \\ =\mathrm{tan\theta }=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$