1.

If `cot theta+ tan theta=m` and `sec theta-cos theta=n` then `(m^2 n)^(2/3)-(mn^2)^(2/3)=`

Answer» `m=((costheta)/(sintheta)+(sintheta)/(costheta))=((cos^(2)theta+sin^(2)theta)/(sinthetacostheta))=(1)/(sinthetacostheta)`
`n=((1)/(costheta)-costheta)=((1-cos^(2)theta)/(costheta))=(sin^(2)theta)/(costheta)`
`therefore m^(2)n=((1)/(sin^(2)theta cos^(2)theta)xx(sin^(2)theta)/(cos^(2)theta))=(1)/(cos^(3)theta)=sec^(3)theta`
`and mn^(2)=((1)/(sintheta costheta)xx(sin^(4)theta)/(cos^(2)theta))=(sin^(3)theta)/(cos^(3)theta)=tan^(3)theta.`
`therefore (m^(2)n)^(2//3)-(mn^(2))^(2//3)=(sec^(3)theta)^(2//3)-(tan^(3)theta)^(2//3)=(sec^(2)theta-tan^(2)theta)=1.`


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