How do you express $sin^2 theta - sec^2 thetacottheta + tan^2 theta$ in terms of $cos theta$ ?

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Answer 1 · 2018-06-30T18:10:04

$(1-cos^4(theta))/cos^2(theta)-1/(cos(theta)sin(theta))$ where

$sin(theta)=pmsqrt(1-cos^2(theta))$

Using

$sin^2(theta)=1-cos^2(theta)$
and
$sec^2(theta)=1/cos^2(theta)$
$cot(theta)=cos(theta)/sin(theta)=cos(theta)/(pmsqrt(1-cos^2(theta))$

$tan^2(theta)=sin^2(theta)/cos^2(theta)=(1-cos^2(theta))/cos^2(theta)$
putting Things together

$1-cos^2(theta)+(1-cos^2(theta))/cos^2(theta)-1/(sin(theta)cos(theta))$

this is equal to

$((1-cos^2(theta))(1+cos^2(theta)))/cos^2(theta)-1/(cos(theta)*sin(theta))$
using that

$a^2-b^2=(a-b)(a+b)$

we get

$(1-cos^4(theta))/cos^2(theta)-1/(cos(theta)sin(theta))$

where

$sin(theta)=pmsqrt(1-cos^2(theta))$

How do you express sin^2 theta - sec^2 thetacottheta + tan^2 theta sin^2 theta - sec^2 thetacottheta + tan^2 theta in terms of cos theta cos theta ?