Question

a cos theta+a sin theta=m,a sin theta-b cos theta=n,prove that a²+b²=m²+n²

Answer 1

Answer:It is given below

Step-by-step explanation:

Given that, aCos theta + bSin theta=m......(1)

And,

aSin theta -bCos theta=n.......(2)

To prove: a^2+b^2= m^2+n^2

Squaring both sides of equation (1) we get

m^2=(aCos theta +bSin theta)^2

=a^2Cos^2 theta+ 2abCos theta Sin theta+b^2Sin^2 theta)

Now squaring both sides of equation (2) we get

n^2=(aSin theta-bCos theta)^2

=a^2Sin^2 theta-2ab Sin theta Coz theta+ b^2cos^2 theta

Now adding these above two equations we get

m^2+n^2=(a^2cos^2 theta+ a^2Sin^2 theta)+(b^2Sin^2 theta+b^2Cos^2 theta)

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

=(a^2)×1+(b^2)×1

=a^2+b^2

Therefore RHS=LHS

Answer 2

Step-by-step explanation:

Hope it will help you