If `a sintheta + bcos theta = c,` then prove that `a costheta -b sintheta = sqrt(a^2+b^2-c^2)`

Explanation

Given that, `asintheta+bcostheta=c`
On squaring both sides,
`(a.sintheta+costheta.b)^(2)=c^(2)`
`rArr a^(2)sin^(2)theta + b^(2)costheta + (2ab)sintheta. costheta=c^(2)` `[therefore(x+y)^(2)=x^(2)+2xy+y^(2)]`
`rArr a^(2)(1-cos^(2)theta) + b^(2)(1-sin^(2)theta)+(2ab)sintheta.costheta=c^(2)` `[therefore sin^(2)theta+ cos^(2)theta=1]`
`rArr a^(2)-a^(2)cos^(2)theta+b^(2)-b^(2)sin^(2)theta+(2ab)sintheta. costheta=c^(2)`
`rArr a^(2)+b^(2)-c^(2)=a^(2)cos^(2)theta+b^(2)sin^(2)theta-(2ab)sintheta.costheta`
`rArr (a^(2)+b^(2)-c^(2)) = (acostheta-bsintheta)^(2)` `[therefore a^(2)+b^(2)-2ab = (a-b)^(2)]`
`rArr (a costheta)-bsintheta = sqrt(a^(2)+b^(2)-c^(2))` Hence proved.