Question

How do you prove this trig identity?

$\frac{{sec}^{2} \theta - 1}{ {sec}^{2} \theta } = {sin}^{2} \theta$

Answer 1

On the left side, multiply numerator and denominator by $\cos^2\theta$:

$\dfrac{\sec^2\theta-1}{\sec^2\theta}=\dfrac{\cos^2\theta(\sec^2\theta-1)}{\cos^2\theta\sec^2\theta}=\dfrac{1-\cos^2\theta}1$

which follows from the fact that $\sec\theta=\dfrac1{\cos\theta}$. Then apply the Pythagorean identity,

$\sin^2\theta+\cos^2\theta=1\implies\dfrac{\sec^2\theta-1}{\sec^2\theta}=\sin^2\theta$.

and we're done.