If a `costheta-bsintheta=c ,` prove that a `sintheta+bcostheta=+-sqrt(a^2+b^2-c^2)`

Explanation

Given, ` a cos theta - b sin theta = c. " "...(i) `
Now, ` (a cos theta - b sin theta)^(2) + (a sin theta + b cos theta)^(2) `
`=a^(2) (cos^(2) theta + sin^(2) theta) + b^(2) (sin^(2) theta + cos ^(2) theta)= (a^(2) +b^(2)) .`
Thus, ` (a cos theta - b sin theta)^(2) + (a sin theta + b cos theta)^(2) = (a^(2) + b^(2)) `
`rArr c^(2) + (a sin theta + b cos theta)^(2) = (a^(2) + b^(2)) `
`rArr (a sin theta + b cos theta)^(2) = (a^(2) + b^(2) - c^(2))`
` rArr (a sin theta + b cos theta ) = pm sqrt(a^(2) + b^(2) - c^(2)). `
Hence, `(a sin theta + b cos theta) = pm sqrt(a^(2) + b^(2) - c^(2)). `