Question

Prove that \( \frac{\sec\theta + \tan\theta}{\sec\theta - \tan\theta} = \left(\frac{1 + \sin\theta}{\cos\theta}\right)^2 \).

Hint:

Quadratic Discriminant: If \( D>0 \), roots are real and distinct; if \( D = 0 \), equal; if \( D<0 \), complex.

Answer 1

Using identities:
\[ \sec\theta + \tan\theta = \frac{1 + \sin\theta}{\cos\theta}, \quad \sec\theta - \tan\theta = \frac{1 - \sin\theta}{\cos\theta} \] \[ \frac{\sec\theta + \tan\theta}{\sec\theta - \tan\theta} = \frac{\frac{1 + \sin\theta}{\cos\theta}}{\frac{1 - \sin\theta}{\cos\theta}} \] \[ = \frac{1 + \sin\theta}{1 - \sin\theta} \] Multiplying numerator and denominator by \( 1 + \sin\theta \):
\[ = \frac{(1 + \sin\theta)^2}{(1 - \sin\theta)(1 + \sin\theta)} \] \[ = \frac{(1 + \sin\theta)^2}{1 - \sin^2\theta} \] Since \( 1 - \sin^2\theta = \cos^2\theta \): \[ = \left(\frac{1 + \sin\theta}{\cos\theta}\right)^2 \] Thus, the identity is proved.
Correct Answer: \( \left( \frac{1 + \sin\theta}{\cos\theta} \right)^2 \)