Answer
$\frac{\cos\alpha}{\sec\alpha}+\frac{\sin\alpha}{\csc\alpha}=\sec^2\alpha-\tan^2\alpha$
Work Step by Step
Simplify the left side:
$\frac{\cos\alpha}{\sec\alpha}+\frac{\sin\alpha}{\csc\alpha}$
$=\frac{\cos\alpha}{\frac{1}{\cos\alpha}}+\frac{\sin\alpha}{\frac{1}{\sin\alpha}}$
$=\frac{\cos\alpha}{\frac{1}{\cos\alpha}}*\frac{\cos\alpha}{\cos\alpha}+\frac{\sin\alpha}{\frac{1}{\sin\alpha}}\frac{\sin\alpha}{\sin\alpha}$
$=\cos^2\alpha+\sin^2\alpha$
$=1$
Simplify the right side:
$\sec^2\alpha-\tan^2\alpha$
$=1$
Since both sides are equal to $1$, they are equal to each other, and the identity is proven.