Question

If aCos theta - bsin theta=x and asin theta + bcos theta = y then prove that a square+ b square = x square + y square pls send answer pls fast​

Answer 1

Answer:-

(Theta is taken as "A")

Given:-

a Cos A - b sin A = x

Squaring both sides we get,

→ (a Cos A - b sin A)² = x²

Similarly,

→ (a Sin A + b Cos A)² = y².

We have to prove:

a² + b² = x² + y²

Putting the values of x² and y² we get,

→ a² + b² = (a Cos A - b Sin A)² + (a Sin A + b Cos A)²

Using the formulae (a + b)² = a² + b² + 2ab and (a - b)² = a² + b² - 2ab in RHS we get,

→ a² + b² = a² cos² A + b² Sin² A - 2ab * Cos A *

Sin A + a² Sin² A + b² Cos² A + 2ab * Cos A * Sin A

→ a² + b² = a² cos² A + a² sin² A + b² sin² A + b² Cos² A

→ a² + b² = a²(Cos² A + Sin² A) + b²(sin² A + cos² A)

Using the identity Cos² A + Sin² A = 1 in RHS we get,

→ a² + b² = a²(1) + b²(1)

→ a² + b² = a² + b²

→ LHS = RHS.

Hence, Proved.

Answer 2

refer to the attachment