Question

if a cos theta minus B sin theta is equal to X, and a sin theta + b cos theta is equal to Y, prove that a square + b square is equal to x squared plus y squared

Answer 1

I hope this is the answer

Answer 2

Solution:

Now the question says$a \cos \theta-b \sin \theta=X$ and the value of $Y={a\sin} \theta+b \cos \theta$

As the question says, to prove$a^{2}+b^{2}=x^{2}+y^{2}$we will put the value of x and y in the equation and square it off, after squaring it off we get the equation:

Squaring the sides we get

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

Subtracting the common figures we get,

=$a \cos ^{2} \theta+b \sin ^{2} \theta+\ {asin}^{2} \theta+b \cos ^{2} \theta$

Taking a and b common we get$\sin ^{2} \theta+\cos ^{2} \theta=1$

$a(1)^{2}+b(1)^{2}$

Therefore, it can be said that the value of$a(1)^{2}+b(1)^{2}=x^{2}+y^{2}$

Hence, Proved.