Question

if a cosA + b sinA =m and a sinA-b cosA =n Prove that a2+b2=m2+n2

Answer 1

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Answer 2

$Given \: (a cos A + b sin A ) = m \: --(1)$

$and \: \: (a sin A - b cos A ) = m \: --(2)$

$(a cos A + b sin A )^{2} + (a sin A - b cos A )^{2} \\= m^{2} + n^{2}$

$\implies (a cosA)^{2} + (b sinA)^{2} + 2 \times (a cos A) \times ( b sin A) \\+ (a sinA)^{2} + (b cos A)^{2} - 2 \times (a sin A) \times ( b cos A) = m^{2} + n^{2}$

$\implies a^{2} cos^{2}A + b^{2} sin^{2} A+ 2 ab sinA cos A\\+ a^{2} sin^{2} A + b^{2} cos^{2} A - 2 ab sin A cos A= m^{2} + n^{2}$

$\implies a^{2}( cos^{2}A + sin^{2} A) + b^{2} (sin^{2} A + cos^{2} A) = m^{2} + n^{2}$

$\implies a^{2} + b^{2} = m^{2} + n^{2}$

$\boxed { \pink { Cos^{2} A + Sin^{2} A = 1 }}$

$Hence \: proved$

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