SinA = 1/x then CosA = ?1. \({\sqrt{x^2 + 1} \over x}\)2. \(x \over \s

1.	SinA = 1/x then CosA = ?1. \({\sqrt{x^2 + 1} \over x}\)2. \(x \over \sqrt{x^2-1}\)3. \({\sqrt{x^2-1} \over x}\)4. \(x \over \sqrt{x^2 + 1}\)
Answer» Correct Answer - Option 3 : \({\sqrt{x^2-1} \over x}\) Given: SinA = 1/x Concept Used: Sinθ = Perpendicular / Hypotenuse Cosθ = Base / Hypotenuse Base2 + Perpendicular2 = Hypotenuse² Calculation: Sinθ = Perpendicular / Hypotenuse SinA = 1/x Comparing get, Perpendicular = 1 and Hypotenuse = x Base = √x² - 1 CosA = Base / Hypotenuse ⇒ CosA = \({\sqrt{x^2-1} \over x}\) ∴ CosA = \({\sqrt{x^2-1} \over x}\).

SinA = 1/x then CosA = ?1. \({\sqrt{x^2 + 1} \over x}\)2. \(x \over \sqrt{x^2-1}\)3. \({\sqrt{x^2-1} \over x}\)4. \(x \over \sqrt{x^2 + 1}\)

Answer» Correct Answer - Option 3 : \({\sqrt{x^2-1} \over x}\)

Given:

SinA = 1/x

Concept Used:

Sinθ = Perpendicular / Hypotenuse

Cosθ = Base / Hypotenuse

Base2 + Perpendicular2 = Hypotenuse²

Calculation:

Sinθ = Perpendicular / Hypotenuse

SinA = 1/x

Comparing get,

Perpendicular = 1 and Hypotenuse = x

Base = √x² - 1

CosA = Base / Hypotenuse

⇒ CosA = \({\sqrt{x^2-1} \over x}\)

∴ CosA = \({\sqrt{x^2-1} \over x}\).