Question

Prove that:
$\frac{\cos x}{1-\sin x}=\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)$

Answer 1

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

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

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