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If cotx(1+sinx)=4m andcotx(1-sinx)=4n , ...

If `cotx(1+sinx)=4m andcotx(1-sinx)=4n ,` prove that `(m^2-n^2)^2=m n` .

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Verified by Experts

Given,
`cotx(1+sinx)=4m and cotx(1-sinx)=4n`
so,
`=>m^2-n^2=(1/16)cot^2x[(1+sinx)^2-(1-sinx)^2]`
`=>m^2-n^2=(1/16)[4sinxcot^2x]`
`=>(m^2-n^2)^2=(sin^2xcot^4x)/(16)`
`=>(m^2-n^2)^2=((sin^2x)/16)((cos^2x)/sin^2x)(cos^2x)`
`=>(m^2-n^2)^2=(cos^4xcos^@x)/(16)----(i)`
...
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