If cot A = k, then sin A is equal to:(presume that A is an acute angle

1.	If cot A = k, then sin A is equal to:(presume that A is an acute angle)1. \(\frac{k^2}{\sqrt {1+k^2}}\)2. \(\frac{1}{k}\)3. \(-\frac{1}{k}\)4. \(\frac{1}{\sqrt {1+k^2}}\)
Answer» Correct Answer - Option 4 : \(\frac{1}{\sqrt {1+k^2}}\) Given: cot A = k Formula: cot A = Base/Perpendicular sin A = Perpendicular/Hypotenuse (Hypotenuse)² = (Base)² + (Perpendicular)² Calculation: cot A = k/1 ⇒ Base/Perpendicular = k/1 ⇒ Base = k and Perpendicular = 1 (Hypotenuse)2 = (Base)2 + (Perpendicular)2 ⇒ (Hypotenuse)2 = k2 + 12 ⇒ Hypotenuse = √(1 + K²) Now, sin A = 1/√(1 + k²)

If cot A = k, then sin A is equal to:(presume that A is an acute angle)1. \(\frac{k^2}{\sqrt {1+k^2}}\)2. \(\frac{1}{k}\)3. \(-\frac{1}{k}\)4. \(\frac{1}{\sqrt {1+k^2}}\)

Answer» Correct Answer - Option 4 : \(\frac{1}{\sqrt {1+k^2}}\)

Given:

cot A = k

Formula:

cot A = Base/Perpendicular

sin A = Perpendicular/Hypotenuse

(Hypotenuse)² = (Base)² + (Perpendicular)²

Calculation:

cot A = k/1

⇒ Base/Perpendicular = k/1

⇒ Base = k and Perpendicular = 1

(Hypotenuse)2 = (Base)2 + (Perpendicular)2

⇒ (Hypotenuse)2 = k2 + 12

⇒ Hypotenuse = √(1 + K²)

Now, sin A = 1/√(1 + k²)