If $$ x = a  sin\theta - b  cos \theta, y = a cos \theta  + b sin \theta $$  , then which of the following is true?

solution

$$ x = a  sin\theta - b  cos \theta$$  {squaring x}

 $$ y = a cos \theta  + b sin \theta $$ {squaring y}

$$ x^2 = a ^2 sin^2\theta + b^2  cos^2 \theta -2absin\theta cos \theta$$

$$ y^2 = a ^2 sin^2\theta + b^2  cos^2 \theta +2absin\theta cos \theta$$

adding both 

we get

$$ x^2 $$+ $$ y^2 $$   = $$a ^2 sin^2\theta + b^2  cos^2 \theta -2absin\theta cos \theta$$ + $$a ^2 sin^2\theta + b^2  cos^2 \theta +2absin\theta cos \theta$$

$$ x^2 $$+ $$ y^2 $$ = $$a ^2 sin^2\theta + b^2  cos^2 \theta + a ^2 sin^2\theta + b^2  cos^2 \theta$$    { $$\because cos^2 \theta +sin^2 \theta = 1$$}

$$ x^2 $$+ $$ y^2 $$ = $$a ^2 (sin^2\theta +  sin^2\theta) +  b^2  (cos^2 \theta +  cos^2 \theta)$$

$$ x^2 $$+ $$ y^2 $$ = $$ a^2 $$+ $$ b^2 $$