What is tan^2theta/(1-sin^4theta)-csc^2theta in terms of non-exponential trigonometric functions?

2 Answers
P dilip_k
Aug 2, 2016

tan^2theta/(1-sin^4theta)-csc^2theta

=sin^2theta/cos^2theta*1/((1-sin^2theta)(1+sin^2theta))-csc^2theta

=sin^2theta/cos^4theta*1/(1+sin^2theta)-csc^2theta

=(8sin^2theta)/(2cos^2theta)^2*1/(2+2sin^2theta)-csc^2theta

=(4(1-cos2theta))/(1+cos2theta)^2*1/(2+1-cos2theta)-csc^2theta

=(8(1-cos2theta))/(2+4cos2theta+2cos^2 2theta)*1/(3-cos2theta)-2/(2sin^2theta)

=(8(1-cos2theta))/(3+4cos2theta+cos4theta)*1/(3-cos2theta)-2/(1-cos2theta)

P dilip_k
Aug 2, 2016

tan^2theta/(1-sin^4theta)-csc^2theta

=sin^2theta/cos^2theta*1/((1-sin^2theta)(1+sin^2theta))-csc^2theta

=sin^2theta/cos^4theta*1/(1+sin^2theta)-csc^2theta

=(8sin^2theta)/(2cos^2theta)^2*1/(2+2sin^2theta)-csc^2theta

=(4(1-cos2theta))/(1+cos2theta)^2*1/(2+1-cos2theta)-csc^2theta

=(8(1-cos2theta))/(2+4cos2theta+2cos^2 2theta)*1/(3-cos2theta)-2/(2sin^2theta)

=(8(1-cos2theta))/(3+4cos2theta+cos4theta)*1/(3-cos2theta)-2/(1-cos2theta)

Related questions
See all questions in Products, Sums, Linear Combinations, and Applications
Impact of this question
2040 views around the world
You can reuse this answer
Creative Commons License