1. **State the problem:** Simplify and verify the equation $\tan^2 x \sin^2 x = \tan^2 x - \sin^2 x$. 2. **Recall definitions and identities:** - $\tan x = \frac{\sin x}{\cos x}$ - $\tan^2 x = \frac{\sin^2 x}{\cos^2 x}$ 3. **Rewrite the left side using $\tan^2 x$:** $$\tan^2 x \sin^2 x = \frac{\sin^2 x}{\cos^2 x} \sin^2 x = \frac{\sin^4 x}{\cos^2 x}$$ 4. **Rewrite the right side:** $$\tan^2 x - \sin^2 x = \frac{\sin^2 x}{\cos^2 x} - \sin^2 x = \frac{\sin^2 x}{\cos^2 x} - \frac{\sin^2 x \cos^2 x}{\cos^2 x}$$ 5. **Combine the right side over common denominator:** $$= \frac{\sin^2 x - \sin^2 x \cos^2 x}{\cos^2 x} = \frac{\sin^2 x (1 - \cos^2 x)}{\cos^2 x}$$ 6. **Use Pythagorean identity $1 - \cos^2 x = \sin^2 x$:** $$= \frac{\sin^2 x \sin^2 x}{\cos^2 x} = \frac{\sin^4 x}{\cos^2 x}$$ 7. **Compare both sides:** Left side = $\frac{\sin^4 x}{\cos^2 x}$, right side = $\frac{\sin^4 x}{\cos^2 x}$, so the equation holds true. **Final answer:** The equation $\tan^2 x \sin^2 x = \tan^2 x - \sin^2 x$ is an identity and is true for all $x$ where $\cos x \neq 0$.