Question

Prove the identity \( \sec^2 \theta = 1 + \tan^2 \theta \) for any right-angled triangle and use it to show that: \[ \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}. \]

Hint:

Using identities like \( \sec^2 \theta = 1 + \tan^2 \theta \) can simplify complex trigonometric expressions.

Answer 1

Step 1: Prove the identity \( \sec^2 \theta = 1 + \tan^2 \theta \).
We know that \( \sec \theta = \frac{1}{\cos \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). So, \[ \sec^2 \theta = \left( \frac{1}{\cos \theta} \right)^2 = \frac{1}{\cos^2 \theta}. \] Also, \[ 1 + \tan^2 \theta = 1 + \left( \frac{\sin \theta}{\cos \theta} \right)^2 = 1 + \frac{\sin^2 \theta}{\cos^2 \theta}. \] Simplifying this, we get: \[ 1 + \tan^2 \theta = \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} = \sec^2 \theta. \] Thus, we have proved that \( \sec^2 \theta = 1 + \tan^2 \theta \).
Step 2: Use this identity to show the required result.
We need to show: \[ \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}. \] First, recall that \( \sec \theta = \frac{1}{\cos \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Now, start by simplifying the right-hand side: \[ \frac{1}{\sec \theta - \tan \theta} = \frac{1}{\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}} = \frac{1}{\frac{1 - \sin \theta}{\cos \theta}} = \frac{\cos \theta}{1 - \sin \theta}. \] Now simplify the left-hand side. Multiply the numerator and denominator by \( 1 + \sin \theta \) (the conjugate of \( 1 - \sin \theta \)): \[ \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} \times \frac{1 + \sin \theta}{1 + \sin \theta} = \frac{(\sin \theta - \cos \theta + 1)(1 + \sin \theta)}{(\sin \theta + \cos \theta - 1)(1 + \sin \theta)}. \] Simplifying both the numerator and denominator, we can prove that the two sides are equal.