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

$\dfrac1s-s=m$

$\dfrac1c-c=n$

In the first equation,

$\dfrac1s-s=\dfrac{1-s^2}s=\dfrac cs=m$

In the second,

$\dfrac1c-c=\dfrac{1-c^2}c=\dfrac sc=n$

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

$\dfrac{c^2}{s^2}+\dfrac{s^2}{c^2}=1$

Now

$\dfrac{c^2}{s^2}+\dfrac{s^2}{c^2}=\dfrac{c^4+s^4}{s^2c^2}=\dfrac{c^4-2s^2c^2+s^4+2s^2c^2}{s^2c^2}=\dfrac{(c^2-s^2)^2}{s^2c^2}+2$

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

$\dfrac{c^2}{s^2}+\dfrac{s^2}{c^2}=\dfrac{(c^2-s^2)^2}{s^2c^2}+2>2$

and cannot possibly be equal to 1.