Answer
$cot^2~\theta+sec^2~\theta = tan^2~\theta+csc^2~\theta$
Work Step by Step
We can prove the identity:
$cot^2~\theta+sec^2~\theta$
$= \frac{cos^2~\theta}{sin^2~\theta}+\frac{1}{cos^2~\theta}$
$= \frac{1-sin^2~\theta}{sin^2~\theta}+\frac{1}{cos^2~\theta}$
$= \frac{1}{sin^2~\theta}-1+\frac{1}{cos^2~\theta}$
$= \frac{1}{sin^2~\theta}-\frac{cos^2~\theta}{cos^2~\theta}+\frac{1}{cos^2~\theta}$
$= \frac{1}{sin^2~\theta}+\frac{1-cos^2~\theta}{cos^2~\theta}$
$= \frac{1}{sin^2~\theta}+\frac{sin^2~\theta}{cos^2~\theta}$
$= \frac{sin^2~\theta}{cos^2~\theta}+\frac{1}{sin^2~\theta}$
$= tan^2~\theta+csc^2~\theta$
