How do you simplify #-1/(cos2theta)+tan2theta# using the double angle identities?

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Answer 1 · 2016-06-11T09:39:37

=#((sintheta-costheta)/(costheta+sintheta))#

#-1/(cos2theta)+tan2theta#

#=-1/(cos2theta)+(sin2theta)/(cos2theta#

#=-(1-sin2theta)/(cos2theta)#

#=-((1-sin2theta)/(cos2theta))#

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

#=-((costheta-sintheta)cancel((costheta-sintheta)))/((costheta+sintheta)cancel((costheta-sintheta)))#

=#-((costheta-sintheta)/(costheta+sintheta))#

=#((sintheta-costheta)/(costheta+sintheta))#