#### Answer

$$\tan^2\alpha\sin^2\alpha=\tan^2\alpha+\cos^2\alpha-1$$
We examine the right side first, which eventually shows that the expression is an identity.

#### Work Step by Step

$$\tan^2\alpha\sin^2\alpha=\tan^2\alpha+\cos^2\alpha-1$$
We examine the right side first.
$$A=\tan^2\alpha+\cos^2\alpha-1$$
- We can rewrite $1$ as follows: $$1=\sin^2\alpha+\cos^2\alpha$$
That makes $A$ into $$A=\tan^2\alpha+\cos^2\alpha-(\sin^2\alpha+\cos^2\alpha)$$
$$A=\tan^2\alpha+\cos^2\alpha-\sin^2\alpha-\cos^2\alpha$$
$$A=\tan^2\alpha-\sin^2\alpha$$
- Now, we would rewrite $\tan^2\alpha$:
$$\tan^2\alpha=\frac{\sin^2\alpha}{\cos^2\alpha}$$
Then $A$ is $$A=\frac{\sin^2\alpha}{\cos^2\alpha}-\sin^2\alpha$$
$$A=\frac{\sin^2\alpha-\sin^2\alpha\cos^2\alpha}{\cos^2\alpha}$$
$$A=\frac{\sin^2\alpha(1-\cos^2\alpha)}{\cos^2\alpha}$$
- Then we rewrite $$1-\cos^2\alpha=\sin^2\alpha$$
$$A=\frac{\sin^2\alpha\sin^2\alpha}{\cos^2\alpha}$$
$$A=\Big(\frac{\sin\alpha}{\cos\alpha}\Big)^2\sin^2\alpha$$
$$A=\tan^2\alpha\sin^2\alpha$$ (since $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$)
2 sides are proved to be equal and the expression is an identity.