Question

If x= a sin theta + b cos theta and, y = a cos theta -b sin theta then find x^2+y^2​

Answer 1

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Answer 2

Answer:

x^2+y^2=(a^2+b^2)

Step-by-step explanation:

x is a sin theta+b cos theta

y is a cos theta- b sin theta

x^2 is a^2 sin^2 theta +b^2 cos^2 theta +(2 ab sin theta cos theta)

y^2 is a^2 cos ^2 theta +b^2 sin^2 theta -(2 ab sin theta cos theta)

x^2+y^2 is a^2 sin^2 theta +b^2 cos^2 theta +(2 ab sin theta cos theta) + a^2 cos ^2 theta +b^2 sin^2 theta -(2 ab sin theta cos theta)

then

=(a^2+b^2)sin^2 theta + (a^2+b^2)cos^2 theta+0

=(a^2+b^2)(sin^2 theta+cos^2 theta)

=(a^2+b^2)

therefore

x^2+y^2=(a^2+b^2)