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

Question

1642 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-31T12:06:43

here is how,

#sintheta-cot^2theta+tan^2theta#

#=sqrt(sin^2theta)-(cos^2theta)/sin^2theta+(sin^2theta)/cos^2theta#

#=sqrt(1-cos^2theta)-cos^2theta/(sqrt(1-cos^2theta))+(sqrt(1-cos^2theta))/cos^2theta#

#= sqrt(1-cos^2theta)+(1-cos^2theta-cos^4theta)/(cos^2thetasqrt(1-cos^2theta))#

#=(cos^2theta(1-cos^2theta)+1-cos^2theta-cos^4theta)/(cos^2thetasqrt(1-cos^2theta))#

#=(cancel(cos^2theta)-cos^4theta+1-cos^4thetacancel(-cos^2theta))/(cos^2thetasqrt(1-cos^2theta)#

#=(1-2cos^4theta)/(sqrt(cos^4theta-cos^6theta))#