a cosA + b sinA=1 a sinA +b cosA=1 PROOVE:  `a^2+b^2=2`

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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.

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