Question

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

Answer 1

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

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

On dividing the numerator and denominator by cos x, we get
$$\begin{gathered} =\frac{1-\tan x}{1+\tan x} \\ =\tan \left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$