a cosA + b sinA=1 a sinA +b cosA=1 PROOVE: `a^2+b^2=2`
The request of the problem is not correct since it is not possible for both provided relations to be valid on the same time, hence, either `a cosA + b sinA = 1` and `a` `sinA - b cosA = 1` , or `cosA - b sinA = 1` and `a sinA + b cosA = 1` .
You need to square the first given relation, such that:
`(acos A + b sin A)^2 = 1^2`
Expanding the square, yields:
`a^2*cos^2 A + 2ab sin A*cos A + b^2 sin^2 A = 1`
You need to square the second modified relation, such that:
`(a sinA - b cosA)^2 = 1^2`
`a^2*sin^2 A - 2ab sin A*cos A + b^2 cos^2 A = 1`
Adding the resulted relations, after squaring process, yields:
`a^2(cos^2 A + sin^2 A) + 0 + b^2(cos^2 A + sin^2 A) = 1 + 1`
Using Pythagorean identity `cos^2 A + sin^2 A = 1` , yields:
`a^2 + b^2 = 2`
Hence, evaluating if the given identity `a^2 + b^2 = 2` holds yields that the statement ` a^2 + b^2 = 2` is not valid, under the original given conditions, but it holds if you'll work with the suggested correction.