Question

(1-sin^2x)/cos^2x How does that equal one? Im doing trigonometric identities and im stuck.

Answer 1

Assuming you mean,\[\frac{1-\sin^2x}{\cos^2x}\]you have recall the fact\[\cos^2x+\sin^2x=1\]so that\[1-\sin^2x=\cos^2x\]hence\[\frac{1-\sin^2x}{\cos^2x}=\frac{\cos^2x}{\cos^2x}=1\]

Answer 2

Consider the fact, that's not solvable, if \[x=(n+0.5)\cdot\pi\] with \[n \in \mathbb{Z}\]