Question

Prove that:

Answer 1

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

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