# Trigonometryprove trig identity (cot x -1)/(cot x +1)=(1-sin 2x)/(cos 2x)

#### karush

##### Well-known member
\begin{align*} \frac{\cot {x}-1}{\cot{x}+1}&=\frac{1-\sin 2x}{\cos 2x}\\ \frac{\cos {x}-\sin x}{\cos{x}+\sin x} \frac{\cos x-\sin x}{\cos x-\sin x}&= \\ \frac{\cos^2x-2\sin x\cos x+\cos^2 x}{\cos^2 x-\sin^2 x} \end{align*}

so far..

#### topsquark

##### Well-known member
MHB Math Helper
\begin{align*} \frac{\cot {x}-1}{\cot{x}+1}&=\frac{1-\sin 2x}{\cos 2x}\\ \frac{\cos {x}-\sin x}{\cos{x}+\sin x} \frac{\cos x-\sin x}{\cos x-\sin x}&= \\ \frac{\cos^2x-2\sin x\cos x+\cos^2 x}{\cos^2 x-\sin^2 x} \end{align*}

so far..
Check your numerator. It should be $$\displaystyle cos^2(x) - 2~sin(x)~cos(x) + sin^2(x)$$.

Otherwise it's good.

-Dan

#### karush

##### Well-known member
\begin{align*} \frac{\cot {x}-1}{\cot{x}+1}&=\frac{1-\sin 2x}{\cos 2x}\\ \frac{\cos {x}-\sin x}{\cos{x}+\sin x} \frac{\cos x-\sin x}{\cos x-\sin x}&= \\ \frac{\cos^2x-2\sin x\cos x+\sin^2 x} {\displaystyle cos^2x- sin^2x}=\\ \frac{1-\sin 2x}{\cos 2x} \end{align*}

hopefully

MHB Math Helper
Yup, you got it.

-Dan