\(\frac{{1 + \sin \theta }}{{\cos \theta }}\) is equal to which of the following (where \(\theta \ne \frac{\pi }{2}\))?

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 12 Dec 2022 Shift 2)

Answer 1

Given:

\(\frac{{1 + \sin \theta }}{{\cos \theta }}\)

Concept used:

1. a² - b² = (a + b)(a - b)

2. sin² θ + cos² θ = 1

Calculation:

\(\frac{{1 + \sin θ }}{{\cos θ }}\)

⇒ \(\frac{(1 + \sin θ)(1 - \sin θ)}{cos θ \times (1 - \sin θ)}\)

⇒ \(\frac{(1 - \sin^2 θ)}{cos θ \times (1 - \sin θ)}\)

⇒ \(\frac{cos^2 θ }{cos θ \times (1 - \sin θ)}\)

⇒ \(\frac{{\cos θ }}{{1 - \sin θ }}\)

∴ \(\frac{{1 + \sin θ }}{{\cos θ }}\) is equal to \(\frac{{\cos θ }}{{1 - \sin θ }}\).