Question

If (cot θ + tan θ) = m and (sec θ – cos θ) = n, prove that `(m^2 n)^(2//3) - (mn^2)^(2//3) = 1`.

Answer 1

We have `(cot theta + tan theta ) = m and ( sec theta - cos theta )=n`

Now, `m^2 n = [(cot theta + tan theta )^2 (sec theta -  cos theta )]`

                  =`[(1/tan theta + tan theta )^2 (1/cos theta- cos theta )]`

                  =`(1+tan^2 theta)^2/tan^2 theta xx ((1-cos^2 theta))/costheta`

                  =`sec^4 theta/tan^2 theta xx sin^2 theta/ cos theta`

                  =`sec ^4 theta /(sin^2 theta/cos^2 theta) xx sin^2 theta / cos theta`

                  =`(cos^2 xxsec^4 theta)/costheta`

                  =`cos theta sec^4 theta`

                 =`1/ sec theta xx sec ^4 theta = sec^3 theta`

∴`(m^2 n)^(2/3) =(sec^3 theta )^(2/3) =  sec^2 theta`

Again, `mn^2 = [(cot theta + tan theta )( sec theta - cos theta )^2 ]`

                      =`[(1/tan theta + tan theta).(1/ cos theta - cos theta)^2]`

                     =`((1+ tan^2 theta))/tan theta xx ((1- cos^2 theta)^2)/cos^2 theta `

                     =`sec^2 theta/tan theta xx sin^4 theta/cos^2 theta`

                    =`sec^2 theta/(sintheta/costheta) xx sin^4 theta/ cos^2 theta`

                    =`(sec^2 xx sin^3 theta)/cos theta`

                     =`1/ cos^2 theta xx sec^3 theta/ cos theta = tan^3 theta `

∴ `(mn^2)^(2/3) = (tan ^3 theta )^(2/3) = tan^2 theta`

Now,` (m^2n)^(2/3) - (mn^2)^(2/3)`

                   =`sec^2 theta - tan^2 theta =1 `

                   =RHS

Hence proved.