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Mathematics

If sin θ + cos θ = 2\sqrt{2} sin(90° - θ), show that cot θ = 2\sqrt{2} + 1.

Trigonometric Identities

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Answer

Given,

    sin θ + cos θ = 2\sqrt{2} sin(90° - θ)
⇒ sin θ + cos θ = 2\sqrt{2} cos θ

Dividing both sides by sin θ

⇒ 1 + cot θ = 2\sqrt{2} cot θ
⇒ 1 = 2\sqrt{2} cot θ - cot θ
⇒ 1 = cot θ(2\sqrt{2} - 1)
⇒ cot θ = 12−1\dfrac{1}{\sqrt{2} - 1}

Rationalizing,

⇒cot θ=12−1×2+12+1⇒cot θ=2+12−1=2+1.⇒ \text{cot θ} = \dfrac{1}{\sqrt{2} - 1} \times \dfrac{\sqrt{2} + 1}{\sqrt{2} + 1} \\[1em] ⇒ \text{cot θ} = \dfrac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1.

Hence, proved that cot θ = 2\sqrt{2} + 1.

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