Prove the following identity tan^2(x)sin^2(x)=tan^2(x)-sin^2(x)

1. Given identity:

\tan^2(x)\sin^2(x)=\tan^2(x)-\sin^2(x)

2. Using the definition of tangent:

\tan(x) = \frac{\sin(x)}{\cos(x)}

3. Substituting $\tan(x)$ in the given equation:

\left(\frac{\sin(x)}{\cos(x)}\right)^2 \sin^2(x) = \left(\frac{\sin(x)}{\cos(x)}\right)^2 - \sin^2(x)

4. Simplifying the left-hand side:

\frac{\sin^2(x) \cdot \sin^2(x)}{\cos^2(x)} = \frac{\sin^4(x)}{\cos^2(x)}

5. Simplifying the right-hand side:

\frac{\sin^2(x)}{\cos^2(x)} - \sin^2(x)

\frac{\sin^2(x)}{\cos^2(x)} - \frac{\sin^2(x) \cos^2(x)}{\cos^2(x)}

\frac{\sin^2(x)}{\cos^2(x)} - \sin^2(x) = \frac{\sin^2(x) - \sin^2(x)\cos^2(x)}{\cos^2(x)}

6. Verifying the equality:

\frac{\sin^2(x)\sin^2(x)}{\cos^2(x)}=\frac{\sin^2(x)^2 - \sin^2(x) \cos^2(x)}{\cos^2(x)}

Finally, we therefore see the truth in the original statement.

\tan^2(x)\sin^2(x)=\tan^2(x)-\sin^2(x)