1. **State the problem:** Prove that $$\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}} = \csc \theta + \cot \theta$$. 2. **Recall identities:** - $$\csc \theta = \frac{1}{\sin \theta}$$ - $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$ - Pythagorean identity: $$\sin^2 \theta + \cos^2 \theta = 1$$ 3. **Start with the left-hand side (L.H.S):** $$\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}$$ 4. **Multiply numerator and denominator inside the square root by $$1 + \cos \theta$$ to rationalize the denominator:** $$\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta} \times \frac{1 + \cos \theta}{1 + \cos \theta}} = \sqrt{\frac{(1 + \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)}}$$ 5. **Simplify the denominator using difference of squares:** $$(1 - \cos \theta)(1 + \cos \theta) = 1 - \cos^2 \theta = \sin^2 \theta$$ 6. **So the expression becomes:** $$\sqrt{\frac{(1 + \cos \theta)^2}{\sin^2 \theta}} = \frac{1 + \cos \theta}{\sin \theta}$$ 7. **Rewrite the numerator over $$\sin \theta$$:** $$\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} = \csc \theta + \cot \theta$$ 8. **Thus, L.H.S = R.H.S, proving the identity.** **Final answer:** $$\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}} = \csc \theta + \cot \theta$$