Identity with secant and tangent

Text

Prove that, wherever the functions are defined, $\sec^{2}x-\tan^{2}x=1$.

Solution

By definition $\sec x=\dfrac{1}{\cos x}$ and $\tan x=\dfrac{\sin x}{\cos x}$. Then $\sec^{2}x-\tan^{2}x=\frac{1}{\cos^{2}x}-\frac{\sin^{2}x}{\cos^{2}x}=\frac{1-\sin^{2}x}{\cos^{2}x}.$ By the fundamental identity $\sin^{2}x+\cos^{2}x=1$ we have $1-\sin^{2}x=\cos^{2}x$, hence $\frac{\cos^{2}x}{\cos^{2}x}=1.$ $\boxed{\;\sec^{2}x-\tan^{2}x=1\;}$