Question

IF TanA= 2ab / a^ 2 -b^ 2 find the value of SinA.

Answer 1

Given

⇒TanA = 2ab/(a² - b²)

To find the value of SinA

We know that

⇒TanA = 2ab/(a² - b²) = Perpendicular/Base

we get

⇒Perpendicular(p) = 2ab , base(b) = (a² - b²) and Hypotenuse  (h) = x

Using Pythagoras theorem

⇒h² = p² + b²

⇒h² = (2ab)² + (a² - b²)²

⇒h² = 4a²b² + a⁴ + b⁴ - 2a²b²

⇒h² = a⁴ + b⁴ + 2a²b²

⇒h = √(a⁴ + b⁴ + 2a²b²)

we get

⇒Perpendicular(p) = 2ab , base(b) = (a² - b²) and Hypotenuse(h) = √(a⁴ + b⁴ + 2a²b²)

We Know that

⇒SinA = Perpendicular(p)/Hypotenuse(h)

Put the value

⇒SinA = 2ab/{√(a⁴ + b⁴ + 2a²b²)}

Answer

⇒SinA = 2ab/{√(a⁴ + b⁴ + 2a²b²)}

Answer 2

Step-by-step explanation:

Given

⇒TanA = 2ab/(a² - b²)

To find the value of SinA

We know that

⇒TanA = 2ab/(a² - b²) = Perpendicular/Base

we get

⇒Perpendicular(p) = 2ab , base(b) = (a² - b²) and Hypotenuse  (h) = x

Using Pythagoras theorem

⇒h² = p² + b²

⇒h² = (2ab)² + (a² - b²)²

⇒h² = 4a²b² + a⁴ + b⁴ - 2a²b²

⇒h² = a⁴ + b⁴ + 2a²b²

⇒h = √(a⁴ + b⁴ + 2a²b²)

we get

⇒Perpendicular(p) = 2ab , base(b) = (a² - b²) and Hypotenuse(h) = √(a⁴ + b⁴ + 2a²b²)

We Know that

⇒SinA = Perpendicular(p)/Hypotenuse(h)

Put the value

⇒SinA = 2ab/{√(a⁴ + b⁴ + 2a²b²)}

Answer

⇒SinA = 2ab/{√(a⁴ + b⁴ + 2a²b²)}

hope this helps you!!

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