## Precalculus (10th Edition)

Work on the left side of the identity using $\tan{\theta}=\frac{\sin\theta}{\cos\theta}$, $\cot\theta=\frac{\cos\theta}{\sin\theta}$, and $\sin^2\theta+\cos^2\theta=1$. Refer to the step-by-step part below for the complete proof.
We have to show that: $\tan^2\theta\cos^2\theta+\cot^2\theta\sin^2\theta=1$ Using $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\cot\theta=\frac{\cos\theta}{\sin\theta}$, the expression above simplifies to: $=\left(\frac{\sin^2\theta}{\cos^2\theta}\right)\cos^2\theta+\left(\frac{\cos^2\theta}{\sin^2\theta}\right)\sin^2\theta$ $=\sin^2\theta+\cos^2\theta$ Since $\sin^2\theta+\cos^2\theta=1$, then the expression above is equivalent to:=1\$ With LHS=RHS, the proof of the identity is complete.