Question

Prove that:

$\frac{\cos \mathrm{\theta }}{1-\sin \mathrm{\theta }}=\tan \left(\frac{\pi }{4}+\frac{\mathrm{\theta }}{2}\right)$

Answer 1

$\mathrm{LHS}=\frac{\mathrm{cos\theta }}{1-\mathrm{sin\theta }}$
$$\begin{gathered} =\frac{{\cos }^2{\displaystyle \frac{\mathrm{\theta }}{2}}-{\sin }^2{\displaystyle \frac{\mathrm{\theta }}{2}}}{{\sin }^2{\displaystyle \frac{\mathrm{\theta }}{2}}+{\cos }^2{\displaystyle \frac{\mathrm{\theta }}{2}}-2\sin {\displaystyle \frac{\mathrm{\theta }}{2}}\times \cos {\displaystyle \frac{\mathrm{\theta }}{2}}}\left[\because \mathrm{cos\theta }={\cos }^2\frac{\mathrm{\theta }}{2}-{\sin }^2\frac{\mathrm{\theta }}{2},\mathrm{sin\theta }=2\sin \frac{\mathrm{\theta }}{2}\cos \frac{\mathrm{\theta }}{2}\mathrm{and}{\sin }^2\frac{\mathrm{\theta }}{2}+{\cos }^2\frac{\mathrm{\theta }}{2}=1\right] \\ =\frac{\left(\cos {\displaystyle \frac{\mathrm{\theta }}{2}}-\sin {\displaystyle \frac{\mathrm{\theta }}{2}}\right)\left(\cos {\displaystyle \frac{\mathrm{\theta }}{2}}+\sin {\displaystyle \frac{\mathrm{\theta }}{2}}\right)}{{\left(\cos {\displaystyle \frac{\mathrm{\theta }}{2}}-\sin {\displaystyle \frac{\mathrm{\theta }}{2}}\right)}^2} \\ =\frac{\cos {\displaystyle \frac{\mathrm{\theta }}{2}}+\sin {\displaystyle \frac{\mathrm{\theta }}{2}}}{\cos {\displaystyle \frac{\mathrm{\theta }}{2}}-\sin {\displaystyle \frac{\mathrm{\theta }}{2}}} \end{gathered}$$

On dividing the numerator and denominator by $\cos \frac{\mathrm{\theta }}{2}$, we get

$$\begin{gathered} =\frac{1+\tan \frac{\mathrm{\theta }}{2}}{1-\tan \frac{\mathrm{\theta }}{2}} \\ =\tan \left(\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\theta }}{2}\right)=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$