If `cos e ctheta-sintheta=ma n dsectheta-costheta=n ,` prove that `(m^(2n))^(2/3)+(m m^2)^(2/3)=1`

Explanation

We have
` m^(2)n = ("cosec" theta - sin theta)^(2)* (sec theta - cos theta) `
` =((1)/(sin theta) - sin theta)^(2) * ((1)/(cos theta) - cos theta) `
`= ((1- sin^(2) theta)^(2) )/(sin^(2)theta) * ((1- cos^(2)theta))/(cos theta) = (cos^(4)theta)/(sin^(2)theta) xx (sin^(2) theta)/(cos theta) = cos^(3) theta `
` therefore (m^(2)n)^(1//3) = cos theta . " "...(i) `
Again, `mn^(2) = ("cosec" theta - sin theta)* (sec theta - cos theta)^(2) `
`= ((1)/(sin theta)-sin theta) * ((1)/(cos theta) - cos theta)^(2) `
`= ((1- sin^(2) theta))/(sin theta) * ((1- cos^(2)theta)^(2))/(cos^(2) theta) `
`= ((cos^(2) theta)/(sin theta) xx (sin^(4) theta)/(cos^(2) theta)) = sin^(3) theta `
` therefore (mn^(2))^(1//3) = sin theta . " "...(ii) `
On squaring (i) and (ii) and adding the results, we get
` (m^(2)n)^(2//3) + (mn^(2))^(2//3) = 1 " "[because cos^(2) theta + sin^(2) theta =1]. `
Hence,`(m^(2)n)^(2//3) + (mn^(2))^(2//3) =1. `