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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`.
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Solution
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.
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