Trig3.3

Prove that $\cot (\frac{\pi}{4}-\theta)=\frac{\cos 2\theta}{1-\sin 2\theta}\,$ . Hence find the value of $\cot 15^\circ\,$

RHS= $\frac{\cos 2\theta}{1-\sin 2\theta}\,$

$\frac{\sin (\frac{\pi}{2}-2\theta)}{1-\cos (\frac{\pi}{2}-2\theta}\,$

$\frac{\sin 2(\frac{\pi}{4}-\theta)}{1-\cos 2(\frac{\pi}{4}-\theta)}\,$

$\frac{2\sin (\frac{\pi}{4}-\theta) \cos (\frac{\pi}{4}-\theta)}{2\sin^2 (\frac{\pi}{4}-\theta)}\,$

$\frac{\cos (\frac{\pi}{4}-\theta)}{\sin (\frac{\pi}{4}-\theta)}=\cot (\frac{\pi}{4}-\theta)\,$ =LHS

Put $\theta=30^\circ\,$

Hence $\cot 15=\frac{\cos 60}{1-\sin 60}=\frac{\frac{1}{2}}{1-\frac{\sqrt{3}}{2}}=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}\,$