1. **State the problem:** Solve and verify the trigonometric identity: $$\frac{1}{\cot^2 x} + \frac{1}{\sin x \cos x} = \sec^2 x$$ 2. **Recall formulas and identities:** - $\cot x = \frac{\cos x}{\sin x}$ - $\sec x = \frac{1}{\cos x}$ - $\cot^2 x = \frac{\cos^2 x}{\sin^2 x}$ - Pythagorean identity: $\sin^2 x + \cos^2 x = 1$ 3. **Rewrite the left side using $\cot^2 x$:** $$\frac{1}{\cot^2 x} = \frac{1}{\frac{\cos^2 x}{\sin^2 x}} = \frac{\sin^2 x}{\cos^2 x}$$ 4. **Rewrite the second term:** $$\frac{1}{\sin x \cos x}$$ 5. **Sum the two terms on the left:** $$\frac{\sin^2 x}{\cos^2 x} + \frac{1}{\sin x \cos x}$$ 6. **Find common denominator $\cos^2 x \sin x$:** $$\frac{\sin^2 x \sin x}{\cos^2 x \sin x} + \frac{\cos^2 x}{\cos^2 x \sin x} = \frac{\sin^3 x + \cos^2 x}{\cos^2 x \sin x}$$ 7. **Simplify numerator if possible:** No direct simplification, so keep as is. 8. **Check if this equals $\sec^2 x = \frac{1}{\cos^2 x}$:** Multiply both sides by $\cos^2 x \sin x$: $$\sin^3 x + \cos^2 x = \frac{\cos^2 x \sin x}{\cos^2 x} = \sin x$$ 9. **Rewrite equation:** $$\sin^3 x + \cos^2 x = \sin x$$ 10. **Rearrange:** $$\sin^3 x - \sin x + \cos^2 x = 0$$ 11. **Use $\cos^2 x = 1 - \sin^2 x$:** $$\sin^3 x - \sin x + 1 - \sin^2 x = 0$$ 12. **Group terms:** $$\sin^3 x - \sin^2 x - \sin x + 1 = 0$$ 13. **Factor by grouping:** $$\sin^2 x (\sin x - 1) - (\sin x - 1) = (\sin x - 1)(\sin^2 x - 1) = 0$$ 14. **Factor $\sin^2 x - 1$ as difference of squares:** $$(\sin x - 1)(\sin x - 1)(\sin x + 1) = 0$$ 15. **Solutions:** $$\sin x = 1 \quad \text{or} \quad \sin x = -1$$ 16. **Check domain restrictions:** $\cot x$ and $\sin x \cos x$ must be defined (nonzero denominators). 17. **Final conclusion:** The identity holds true for values of $x$ where $\sin x = \pm 1$ and denominators are defined. **Answer:** The equation is true for $x$ such that $\sin x = 1$ or $\sin x = -1$ with domain restrictions respected.