Question

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

Answer 1

Step 1: Expanding the left-hand side expression:

We are given the expression \( (\cot \theta - \csc \theta)^2 \) and asked to prove that it equals \( \frac{1 - \cos \theta}{1 + \cos \theta} \).
First, we expand the left-hand side expression using the identity \( (a - b)^2 = a^2 - 2ab + b^2 \):
\[ (\cot \theta - \csc \theta)^2 = \cot^2 \theta - 2 \cot \theta \csc \theta + \csc^2 \theta \] We will now simplify each of the terms.

Step 2: Simplifying the terms:

We know the following standard trigonometric identities:
\[ \cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \csc \theta = \frac{1}{\sin \theta} \] Substituting these into the expanded expression:
1. \( \cot^2 \theta = \left( \frac{\cos \theta}{\sin \theta} \right)^2 = \frac{\cos^2 \theta}{\sin^2 \theta} \)
2. \( \csc^2 \theta = \left( \frac{1}{\sin \theta} \right)^2 = \frac{1}{\sin^2 \theta} \)
3. \( \cot \theta \csc \theta = \frac{\cos \theta}{\sin \theta} \times \frac{1}{\sin \theta} = \frac{\cos \theta}{\sin^2 \theta} \)
Now substitute these into the expanded expression:
\[ (\cot \theta - \csc \theta)^2 = \frac{\cos^2 \theta}{\sin^2 \theta} - 2 \times \frac{\cos \theta}{\sin^2 \theta} + \frac{1}{\sin^2 \theta} \] Simplify the terms by factoring out \( \frac{1}{\sin^2 \theta} \):
\[ = \frac{1}{\sin^2 \theta} \left( \cos^2 \theta - 2 \cos \theta + 1 \right) \] The expression inside the parentheses is a perfect square:
\[ \cos^2 \theta - 2 \cos \theta + 1 = (\cos \theta - 1)^2 \] Thus, the left-hand side becomes:
\[ (\cot \theta - \csc \theta)^2 = \frac{(\cos \theta - 1)^2}{\sin^2 \theta} \]

Step 3: Simplifying the right-hand side expression:

Now, let's simplify the right-hand side expression \( \frac{1 - \cos \theta}{1 + \cos \theta} \). We can multiply both the numerator and denominator by \( 1 - \cos \theta \) to simplify the expression:
\[ \frac{1 - \cos \theta}{1 + \cos \theta} \times \frac{1 - \cos \theta}{1 - \cos \theta} = \frac{(1 - \cos \theta)^2}{(1 + \cos \theta)(1 - \cos \theta)} \] Using the difference of squares identity, \( (a - b)(a + b) = a^2 - b^2 \), we get:
\[ = \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} \] Since \( 1 - \cos^2 \theta = \sin^2 \theta \), we have:
\[ = \frac{(1 - \cos \theta)^2}{\sin^2 \theta} \]

Step 4: Conclusion:

We can now see that both the left-hand side and the right-hand side are identical:
\[ (\cot \theta - \csc \theta)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta} = \frac{1 - \cos \theta}{1 + \cos \theta} \] Thus, we have proven that:
\[ (\cot \theta - \csc \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta} \]