Question

sin theta+sin 2theta/1+cos theta+cos 2theta

Answer 1

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