Question

Prove that \[ \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \csc\theta - \cot\theta \]

Hint:

When proving trigonometric identities involving square roots, rationalize the fraction or use half-angle formulas to simplify.

Answer 1

Starting with the Left-Hand Side (LHS): \[ \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} \] Multiply numerator and denominator by $(1-\cos\theta)$: \[ = \sqrt{\frac{(1-\cos\theta)^2}{(1+\cos\theta)(1-\cos\theta)}} \] \[ = \sqrt{\frac{(1-\cos\theta)^2}{1-\cos^2\theta}} \] \[ = \sqrt{\frac{(1-\cos\theta)^2}{\sin^2\theta}} \] \[ = \frac{1-\cos\theta}{\sin\theta} \] Now simplify: \[ \frac{1-\cos\theta}{\sin\theta} = \frac{1}{\sin\theta} - \frac{\cos\theta}{\sin\theta} \] \[ = \csc\theta - \cot\theta \] Hence, \[ \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \csc\theta - \cot\theta \] \[ \boxed{\text{Proved}} \]