Question

Cosec theta -sin theta =m cube sec theta - cos theta = n cube prove that m^4n^2+n^4m^2=1

Answer 1

This is false. (Is there any more information? Are you sure there is no typo?)

Let
`s=\sin\theta` and
`c=\cos\theta`. Then

`\frac1s-s=m`

`\frac1c-c=n`

In the first equation,

`\frac1s-s=\frac{1-s^2}s=\frac cs=m`

In the second,

`\frac1c-c=\frac{1-c^2}c=\frac sc=n`

So
`mn=1`, and

`m^4n^2+n^4m^2=m^2n^2(m^2+n^2)=m^2+n^2`

so we're left to prove that

`(c^2)/(s^2)+(s^2)/(c^2)=1`

Now

`(c^2)/(s^2)+(s^2)/(c^2)=(c^4+s^4)/(s^2c^2)=(c^4-2s^2c^2+s^4+2s^2c^2)/(s^2c^2)=((c^2-s^2)^2)/(s^2c^2)+2`

and both
`(c^2-s^2)^2` and
`s^2c^2` are non-negative, so

`(c^2)/(s^2)+(s^2)/(c^2)=((c^2-s^2)^2)/(s^2c^2)+2>2`

and cannot possibly be equal to 1.