Question

Prove that $cot (\frac{\pi}{4} - \theta) = \frac{cos 2\theta}{1 - sin 2\theta}$ and hence find the value of cot 15°.

Answer 1

We know that

$Cot(\alpha-\beta)=\frac{Cot\alpha.Cot\beta+1}{Cot\beta-Cot\alpha}\\\\Cot(45-\theta)=\frac{Cot45.Cot\theta+1}{Cot\theta-Cot45}\\\\ \\Cot(45-\theta)=\frac{1.Cot\theta+1}{Cot\theta-1} \\Cot(45-\theta)=\frac{Cos\theta/Sin\theta+1}{Cos\theta/Sin\theta-1}\\ Cot(45-\theta)=\frac{(Cos\theta+Sin\theta)/Sin\theta}{(Cos\theta-Sin\theta)/Sin\theta}\\\\Cot(45-\theta)=\frac{Cos\theta+Sin\theta}{Cos\theta-Sin\theta}$

$Cot(45-\theta)=\frac{(Cos\theta+Sin\theta)(Cos\theta-Sin\theta)}{(Cos\theta-Sin\theta)(Cos\theta-Sin\theta)}\\\\Cot(45-\theta)=\frac{Cos^{2}\theta-Sin^{2}\theta}{Sin^{2}\theta+Cos^{2}\theta-2Sin\theta.Cos\theta}\\\\Cot(45-\theta)=\frac{Cos2\theta}{1-Sin2\theta}$

Now

⇒     Cot 15=(Cot 45 - Cot 30)

Using same formula, we get

$Cot(\alpha-\beta)=\frac{Cot\alpha.Cot\beta+1}{Cot\beta-Cot\alpha}\\\\Cot15=Cot(45-30)=\frac{Cot45.Cot30+1}{Cot30-Cot45}\\\\Cot15=\frac{1\sqrt{3}+1}{\sqrt{3} -1}\\\\Cot15=\frac{\sqrt{3}+1}{\sqrt{3} -1}$

Answer 2

Answer:

Step-by-step explanation: