Question

If xcos θ -ysin θ = a, xsin θ + ycos θ = b, prove that x²+y²=a²+b².

Answer 1

Given :-

xcos Ø - ysinØ = a ----(1)

xsin Ø + ycos Ø = b -----(2)

To Prove :- x² + y² = a² + b²

Squaring and adding both the equations :-

(xcos Ø - ysin Ø)² + (xsin Ø + ycos Ø)² = a² + b²

x²cos²Ø + y²sin²Ø -2xysinØ.cosØ + x²sin²Ø + y²cos²Ø + 2xysinØ.cosØ = a² + b²

x²cos²Ø + x²sin²Ø + y²cos²Ø + y²sin²Ø = a² + b²

x²(cos²Ø+sin²Ø) + y²(cos²Ø+sin²Ø) = a² + b²

As we know that :- sin²Ø + cos²Ø = 1

x²(1) + y²(1) = a² + b²

Therefore,

x² + y² = a² + b²

Proved........

Answer 2

Answer:

given: x cos x - y sin x = a, x sin x + y cos x = b

to prove x² + y² = a² + b²

Step-by-step explanation:

let's take RHS and prove it equal to LHS

RHS => a² + b² =>

a² => (x cos x - y sin x)²

=> x² cos²x + y² sin²x - 2xy cos x sin x.............1

b² => (x sin x + y cos x)²

=> x² sin²x + y² cos²x + 2xy cos x sin x.....,......2

equation 1 + 2

=> x²cos²x+y²sin²x-2xy cos x sin x + x²sin²x+y²cos²x + 2xy cos x sin x

=> x²cos²x + y²sin²x + x²sin²x + y²cos²x

[ 2xy sin x cos x cancels out ]

=> x² [ sin²x + cos²x ] + y² [ sin²x + cos²x ]

=> x² [1] + y² [1 ] [ sin²x + cos²x = 1 ]

=> x² + y² = LHS Hence proved