Question

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

Answer 1

Step 1: Expand the left-hand side of the equation:
We start with the left-hand side of the given equation:
\[ (\cot \theta - \csc \theta)^2 \] First, recall the trigonometric identities for $\cot \theta$ and $\csc \theta$:
\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \quad \text{and} \quad \csc \theta = \frac{1}{\sin \theta} \] Now, substitute these into the expression: \[ (\cot \theta - \csc \theta)^2 = \left( \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta} \right)^2 \] Factor out $\frac{1}{\sin \theta}$: \[ = \left( \frac{\cos \theta - 1}{\sin \theta} \right)^2 \] Now, square both the numerator and the denominator: \[ = \frac{(\cos \theta - 1)^2}{\sin^2 \theta} \] Step 2: Express the right-hand side of the equation:
The right-hand side of the given equation is: \[ \frac{1 - \cos \theta}{1 + \cos \theta} \] Step 3: Manipulate the expression to match both sides:
We will work with the right-hand side. To simplify, multiply both the numerator and the denominator by $(1 - \cos \theta)$: \[ \frac{1 - \cos \theta}{1 + \cos \theta} \times \frac{1 - \cos \theta}{1 - \cos \theta} = \frac{(1 - \cos \theta)^2}{(1 + \cos \theta)(1 - \cos \theta)} \] Now, apply the difference of squares formula in the denominator: \[ (1 + \cos \theta)(1 - \cos \theta) = 1^2 - (\cos \theta)^2 = 1 - \cos^2 \theta \] Using the Pythagorean identity, we know that: \[ \sin^2 \theta = 1 - \cos^2 \theta \] Thus, the denominator becomes $\sin^2 \theta$. Therefore, we have: \[ \frac{(1 - \cos \theta)^2}{\sin^2 \theta} \] Step 4: Conclusion:
We now observe that both the left-hand side and the right-hand side of the equation are the same: \[ \frac{(\cos \theta - 1)^2}{\sin^2 \theta} = \frac{(1 - \cos \theta)^2}{\sin^2 \theta} \] Thus, we have proven that: \[ (\cot \theta - \csc \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta} \]