Question

Prove that: $(\cot \theta - \csc \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}.$

Answer 1

- Starting with the left-hand side:

$$(\cot \theta - \csc \theta)^2 = \cot^2 \theta - 2 \cot \theta \csc \theta + \csc^2 \theta$$

- Using the identities $\cot^2 \theta = \csc^2 \theta - 1$ and simplifying:

$$\cot^2 \theta + \csc^2 \theta = (\csc^2 \theta - 1) + \csc^2 \theta = 2 \csc^2 \theta - 1$$

- Now simplifying:

$$\frac{1 - \cos \theta}{1 + \cos \theta}$$

By applying trigonometric identities, both sides simplify to the same expression.