1. **State the problem:** We want to prove the identity $$\frac{1-\sin\theta}{\cos\theta} = 3\cot\theta$$ 2. **Recall definitions:** - Recall that \(\cot\theta = \frac{\cos\theta}{\sin\theta}\). 3. **Manipulate the left-hand side (LHS):** Start with $$\frac{1 - \sin\theta}{\cos\theta}$$ 4. **Check if the right-hand side (RHS) can be rewritten:** Given RHS is $$3\cot\theta = 3\frac{\cos\theta}{\sin\theta} = \frac{3\cos\theta}{\sin\theta}$$ 5. **Compare LHS and RHS:** - To prove equality, cross-multiply or simplify. Multiply both sides by \(\cos\theta\sin\theta\): $$ (1 - \sin\theta) \sin\theta = 3\cos^2\theta $$ 6. **Expand the left side:** $$\sin\theta - \sin^2\theta = 3\cos^2\theta$$ 7. **Use Pythagorean identity:** Recall that $$\sin^2\theta + \cos^2\theta = 1 \implies \cos^2\theta = 1 - \sin^2\theta$$ 8. **Substitute into equation:** $$\sin\theta - \sin^2\theta = 3(1 - \sin^2\theta)$$ Expand the right side: $$\sin\theta - \sin^2\theta = 3 - 3\sin^2\theta$$ 9. **Bring all terms to one side:** $$\sin\theta - \sin^2\theta - 3 + 3\sin^2\theta = 0$$ Simplify: $$\sin\theta + 2\sin^2\theta - 3 = 0$$ 10. **Check if this holds for all \(\theta\):** This is not an identity true for all \(\theta\). Since the equation does not simplify to a tautology, the original statement is false in general. **Final conclusion:** The given equation $$\frac{1 - \sin\theta}{\cos\theta} = 3\cot\theta$$ is not an identity true for all \(\theta\). It is therefore not generally true. **If this was a problem to prove, please check the original expression as the equality does not hold as stated.**