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

Question

2915 views around the world

You can reuse this answer
Creative Commons License

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))#