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

Question

asked Jan 17, 2020 32.3k views

1 Answer

Xiaoyu · Answer 1 · 2020-01-24T03:55:10+0000

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.