If $x cos A + y sin A = m$ and $x sin A - y cos A = n$ then prove that : $x^{2} + y^{2} = m^{2} + n^{2}$

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Solution

Taking RHS,
$m^{2} + n^{2}$
$= (x c o s A + y s i n A)^{2} + (x s i n A - y c o s A)^{2}$
$= x^{2} c o s^{2} A + y^{2} s i n^{2} A + 2 x y c o s A s i n A + x^{2} s i n^{2} A + y^{2} c o s^{2} A - 2 x y s i n A c o s A$
$= x^{2} (c o s^{2} A + s i n^{2} A) + y^{2} (s i n^{2} + c o s^{2} A)$
$= x^{2} + y^{2}$ [Since, $c o s^{2} A + s i n^{2} A = 1$ ]
$=$ LHS

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