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)#

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