Question

Prove that \(\sqrt{\frac{1+\cos\theta}{1-\cos\theta}} = \frac{1+\cos\theta}{\sin\theta}\)

Hint:

Rationalization is a powerful tool for simplifying expressions with square roots in the denominator. Multiplying by the conjugate of the denominator often helps eliminate the square root by creating a difference of squares.

Answer 1

Step 1: Understanding the Concept:
To prove this trigonometric identity, we will work with the Left Hand Side (LHS). The standard technique for expressions involving square roots of fractions like this is to rationalize the denominator inside the square root.

Step 2: Key Formula or Approach:
We will use the algebraic identity \((a-b)(a+b) = a^2 - b^2\) and the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\), which implies \(1 - \cos^2\theta = \sin^2\theta\).

Step 3: Detailed Explanation:
Let's start with the LHS: \[ \text{LHS} = \sqrt{\frac{1+\cos\theta}{1-\cos\theta}} \] Multiply the numerator and the denominator inside the square root by \((1+\cos\theta)\) to rationalize the denominator: \[ \text{LHS} = \sqrt{\frac{(1+\cos\theta)(1+\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}} \] \[ \text{LHS} = \sqrt{\frac{(1+\cos\theta)^2}{1^2 - \cos^2\theta}} \] Using the identity \(1 - \cos^2\theta = \sin^2\theta\): \[ \text{LHS} = \sqrt{\frac{(1+\cos\theta)^2}{\sin^2\theta}} \] Now, we can take the square root of the numerator and the denominator: \[ \text{LHS} = \frac{1+\cos\theta}{\sin\theta} \] This is equal to the Right Hand Side (RHS).

Step 4: Final Answer:
Since LHS = RHS, the identity is proved.