tanα = a/b then the value of Sinα + Cosα is:1. \(\sqrt{a^2 + b^2} \

1.	tanα = a/b then the value of Sinα + Cosα is:1. \(\sqrt{a^2 + b^2} \over {a + b}\)2. \({a + b} \over \sqrt{a^2 + b^2}\)3. \(\sqrt{a^2 - b^2} \over {a + b}\)4. \({a + b} \over \sqrt{a^2 - b^2}\)
Answer» Correct Answer - Option 2 : \({a + b} \over \sqrt{a^2 + b^2}\) Given: tanα = a/b Concept Used: tanθ = Perpendicular/Base Sinθ = Perpendicular/Hypotenuse Cosθ = Base/Hypotenuse Base² + Perpendicular² = Hypotenuse² Calculation: tanα = a/b and tanα = Perpendicular/Base Comparing above two get, Perpendicular = a and Base = b Hypotenuse = √a² + b² Sinα = Perpendicular/Hypotenuse ⇒ Sinθ = a/√a² + b² Cosα = Base/Hypotenuse ⇒ Cosα = b/√a2 + b2 Sinα + Cosα = (a/√a2 + b²) + (b/√a2 + b²) ⇒ Sinα + Cosα = \({a + b} \over \sqrt{a^2 + b^2}\) ∴ Sinα + Cosα = \({a + b} \over \sqrt{a^2 + b^2}\)

tanα = a/b then the value of Sinα + Cosα is:1. \(\sqrt{a^2 + b^2} \over {a + b}\)2. \({a + b} \over \sqrt{a^2 + b^2}\)3. \(\sqrt{a^2 - b^2} \over {a + b}\)4. \({a + b} \over \sqrt{a^2 - b^2}\)

Answer» Correct Answer - Option 2 : \({a + b} \over \sqrt{a^2 + b^2}\)

Given:

tanα = a/b

Concept Used:

tanθ = Perpendicular/Base

Sinθ = Perpendicular/Hypotenuse

Cosθ = Base/Hypotenuse

Base² + Perpendicular² = Hypotenuse²

Calculation:

tanα = a/b

and tanα = Perpendicular/Base

Comparing above two get,

Perpendicular = a and Base = b

Hypotenuse = √a² + b²

Sinα = Perpendicular/Hypotenuse

⇒ Sinθ = a/√a² + b²

Cosα = Base/Hypotenuse

⇒ Cosα = b/√a2 + b2

Sinα + Cosα = (a/√a2 + b²) + (b/√a2 + b²)

⇒ Sinα + Cosα = \({a + b} \over \sqrt{a^2 + b^2}\)

∴ Sinα + Cosα = \({a + b} \over \sqrt{a^2 + b^2}\)