Question

If in triangle ABC right angled at B sinC=a^2-b^2÷a^2+b^2,then cosC=?

Answer 1

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Answer 2

Answer:

The correct answer to this question is cos c = 2ab / (a² + b²)

Step-by-step explanation:

Given, ΔABC

right angles at B, ∠B = 90°

sin c = (a² - b²) / (a² + b²)

cos c =?

Now,

sin c = (a² - b²) / (a² + b²)

AB / AC = (a² - b²) / (a² + b²)

AB = (a² - b²) and AC = (a² + b²)

BC =?

Using Pythagoras theorem:

AC² = AB² + BC²

BC² = AC² - AB²

      = (a² + b²)² - (a² - b²)²

      = a⁴ + b⁴ + 2a²b² - (a⁴ + b⁴ - 2a²b²)

      = a⁴ + b⁴ + 2a²b² - a⁴ - b⁴ + 2a²b²

      = 4a²b²

∴ BC² = 4a²b²

BC = 2ab

cos c = adjacent side / hypotenuse

cos c = BC / AC

cos c = 2ab / (a² + b²)

Hence, cos c = 2ab / (a² + b²)

Click here for more about Pythagoras theorem:

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