If sin θ + cos θ = \(\sqrt5\) sin(90 - θ), find the value of cot θ.

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 02 Dec 2022 Shift 2)

Answer 1

Given:

sin θ + cos θ = \(\sqrt5\) sin(90 - θ)

Concept used:

1. cot θ = cos θ ÷ sin θ

2. a² - b² = (a + b)(a - b)

Calculation:

sin θ + cos θ = \(\sqrt5\) sin(90 - θ)

⇒ sin θ + cos θ = \(\sqrt5\) cosθ

⇒ cos θ(\(\sqrt5\) - 1) = sinθ

⇒ cot θ = \(\frac {1}{\sqrt5 - 1}\)

⇒ cot θ = \(\frac {\sqrt5 + 1}{(\sqrt5 - 1)(\sqrt5 + 1)}\)

⇒ cot θ = \(\frac {\sqrt5 + 1}{5 - 1}\)

⇒ cot θ = \(\frac {\sqrt5 + 1}{4}\)

∴ The value of cot θ is \(\frac {\sqrt5 + 1}{4}\).