Answer
$$\sin^2\alpha\sec^2\alpha+\sin^2\alpha\csc^2\alpha=\sec^2\alpha$$
The expression has been proved to be an identity.
Work Step by Step
$$\sin^2\alpha\sec^2\alpha+\sin^2\alpha\csc^2\alpha=\sec^2\alpha$$
The left is more complex and we need to deal with it first.
$$A=\sin^2\alpha\sec^2\alpha+\sin^2\alpha\csc^2\alpha$$
- Using the Reciprocal Identities:
$$\sec\alpha=\frac{1}{\cos\alpha}\hspace{1.5cm}\csc\alpha=\frac{1}{\sin\alpha}$$
Therefore $A$ would be
$$A=\sin^2\alpha\times\frac{1}{\cos^2\alpha}+\sin^2\alpha\times\frac{1}{\sin^2\alpha}$$
$$A=\frac{\sin^2\alpha}{\cos^2\alpha}+1$$
Also, from Quotient Identity:
$$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$$
Therefore, $$A=\tan^2\alpha+1$$
Finally, from Pythagorean Identity: $$\tan^2\alpha+1=\sec^2\alpha$$
So, $$A=\sec^2\alpha$$
The expression is hence an identity.
