Question

if sin theta + cos theta = a, sin theta - cos theta = b prove that a2 + b2 = 2​

Answer 1

$Given \: ( sin \theta + cos \theta ) = a \: --(1)$

$and \: (sin \theta - cos \theta ) = b \: --(2)$

$( sin \theta + cos \theta )^{2} + ( sin \theta - cos \theta )^{2} = a^{2} + b^{2}$

$\implies 2( sin^{2} \theta + cos^{2} \theta ) = a^{2} + b^{2}$

$\boxed{ \blue{ \because (x+y)^{2}+(x-y)^{2} = 2(x^{2} +y^{2}) }}$

$\implies 2 \times 1 = a^{2} + b^{2}$

$\boxed{ \green { \because sin^{2} \theta + cos^{2} \theta = 1 }}$

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

$Hence \:proved$

•••♪