Question

in sin A =a-b/a+b find tan A​

Answer 1

✠ Question : If sin A = (a - b)/(a + b). Find tan A

✠ Given : sin A = (a - b)/(a + b)

✠ To Find : Value of tan A

✠ Answer : tan A = (a - b)/(2√(ab))

✠ Explanation : ⤵️ (See Down)

________________________________________

We Know : sin A = Perpendicular/Hypotenuse

We are asked to Find Value of : tan A

Assume a Triangle ABC, Right angled at B , Where :

• Perpendicular = AB
• Base = BC
• Hypotenuse = AC

We also know : tan A = Perpendicular/Base

sin A = Perpendicular/Hypotenuse

⇒ (a - b)/(a + b) = Perpendicular/Hypotenuse

Comparing this , We get :

• Perpendicular = (a - b)
• Hypotenuse = (a + b)

• To Find tan A, we need to have values of Perpendicular and Base, We have Perpendicular Value but we need to find the value of Base.

To Find Base, Use Pythagoras Theorem :

• In a right angled triangle, the square of the hypotenuse is equal to  the sum of the squares of the other two sides.

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

⇒ (a + b)² = (a - b)² + (Base)²

Using Algebraic Identities

⇒ a² + b² + 2ab = a² + b² - 2ab + (Base)²

Adding 2ab to both Sides

⇒ a² + b² + 2ab + 2ab = a² + b² - 2ab + (Base)² + 2ab

⇒ a² + b² + 4ab = a² + b² + (Base)²

Subtracting a² from both Sides

⇒ a² + b² + 4ab - a² = a² + b² + (Base)² - a²

⇒ b² + 4ab  = b² + (Base)²

Subtracting b² from both Sides

⇒ b² + 4ab - b² = b² + (Base)² - b²

⇒ 4ab = (Base)²

Taking Root on Both Sides

⇒ √(4ab) = √(Base)²

⇒ 2√(ab) = Base

Switch Sides

⇒ Base = 2√(ab)

• Now solve for tan A

tan A = Perpendicular/Base

⇒ tan A = (a - b)/(2√(ab))

________________________________________