Originally Posted by

**Siron** You can also prove the identity by using:

$\displaystyle \frac{A}{B }=\frac{C}{D} \Leftrightarrow A\cdot D=B \cdot C$

That means we have to prove:

$\displaystyle \cot^2(x)[\sec(x)-1][1+\sec(x)]=\sec^2(x)[1-\sin(x)][1+\sin(x)]$

$\displaystyle \Leftrightarrow \cot^2(x)[\sec^2(x)-1]=\sec^2(x)[1-\sin^2(x)]$

$\displaystyle \Leftrightarrow \cot^2(x)\left[\frac{1}{\cos^2(x)}-1\right]=\sec^2(x)\cos^2(x)$

$\displaystyle \Leftrightarrow \cot^2(x)\left[\frac{1-\cos^2(x)}{\cos^2(x)}\right]=\frac{\cos^2(x)}{\cos^2(x)}$

$\displaystyle \Leftrightarrow \cot^2(x)\frac{\sin^2(x)}{\cos^2(x)}=1$

$\displaystyle \Leftrightarrow \cot^2(x)\tan^2(x)=1$

$\displaystyle \Leftrightarrow 1=1$