If x¹=xcosA+ysinA and y¹=xsinA-ycosA, show that (x¹)²+(y¹)²=x²+y²​

Expanding each square on the left side, you have

(x cos(A) + y sin(A))² = x² cos²(A) + 2xy cos(A) sin(A) + y² sin²(A)

(x sin(A) - y cos(A))² = x² sin²(A) - 2xy sin(A) cos(A) + y² cos²(A)

so that adding them together eliminates the identical middle terms and reduces to the sum to

x² cos²(A) + y² sin²(A) + x² sin²(A) + y² cos²(A)

Collecting terms to factorize gives us

(y² + x²) sin²(A) + (x² + y²) cos²(A)

(x² + y²) (sin²(A) + cos²(A))

and sin²(A) + cos²(A) = 1 for any A, so we end up with

x² + y²

as required.

assuming you mean addition: addition and subtraction are reverse operations.

step-by-step explanation: