If `tan theta +sin theta =m` and `tan theta -sin theta =n` then prove `m^2-n^2=4sqrt(mn)`

Explanation

We have
` LHS = (m^(2) - n^(2)) `
` = (tan theta + sin theta)^(2) - (tan theta - sin theta)^(2)= 4 tan theta sin theta `
`[ because (a+b)^(2) - (a-b)^(2)= 4ab]. `
`RHS = 4 sqrt(mn) = 4 sqrt((tan theta + sin theta)(tan theta - sin theta)) `
` = 4 sqrt((tan^(2)theta - sin^(2)theta))= 4 * sqrt (((sin^(2)theta)/(cos^(2)theta) - sin^(2)theta)) `
`= 4 * (sqrt(sin^(2)theta - sin^(2)theta cos^(2)theta))/(cos theta) = 4 * (sin theta)/(cos theta) * sqrt(1- cos^(2)theta)`
` = 4 tan theta * sqrt (sin^(2)theta)= 4 tan theta sin theta . `
Thus, LHS = RHS.
Hence, `(m^(2) - n^(2)) = 4sqrt(mn) . `