Answer
$$\frac{\cos\alpha}{\sec\alpha}+\frac{\sin\alpha}{\csc\alpha}=\sec^2\alpha-\tan^2\alpha$$
The equation is verified to be an identity.
Work Step by Step
$$\frac{\cos\alpha}{\sec\alpha}+\frac{\sin\alpha}{\csc\alpha}=\sec^2\alpha-\tan^2\alpha$$
We would try transforming the left side first, using $\sec\alpha=\frac{1}{\cos\alpha}$ and $\csc\alpha=\frac{1}{\sin\alpha}$ $$A=\frac{\cos\alpha}{\sec\alpha}+\frac{\sin\alpha}{\csc\alpha}$$ $$A=\frac{\cos\alpha}{\frac{1}{\cos\alpha}}+\frac{\sin\alpha}{\frac{1}{\sin\alpha}}$$ $$A=\cos^2\alpha+\sin^2\alpha$$ $$A=1$$
($\sin^2\alpha+\cos^2\alpha$ is an identity and equals $1$)
Now we transform the right side to see if it also equals $1$. We would use the identities $\sec\alpha=\frac{1}{\cos\alpha}$ and $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$ $$B=\sec^2\alpha-\tan^2\alpha$$ $$B=\frac{1}{\cos^2\alpha}-\frac{\sin^2\alpha}{\cos^2\alpha}$$ $$B=\frac{1-\sin^2\alpha}{\cos^2\alpha}$$
Remember that we have an identity $\sin^2\alpha+\cos^2\alpha=1$, which means $1-\sin^2\alpha=\cos^2\alpha$. Therefore, $$B=\frac{\cos^2\alpha}{\cos^2\alpha}$$ $$B=1$$
Both left side and right side are equal $1$, that means they are equal with each other. $$\frac{\cos\alpha}{\sec\alpha}+\frac{\sin\alpha}{\csc\alpha}=\sec^2\alpha-\tan^2\alpha$$
The equation is hence verified to be an identity.
