What is #sin^2theta/(1-tantheta) # in terms of #costheta#?

Question

1142 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-03T05:43:57

#costheta((1−cos^2theta))/(costheta-sqrt(1−cos^2theta))#

#sin^2theta/(1−tantheta)# can be written in term of #costheta#, using identities

#sin^2theta=1−cos^2theta# i.e. #sintheta=sqrt(1−cos^2theta)#

and #tantheta=sintheta/costheta#

As such #sin^2theta/(1−tantheta)#

= #(1−cos^2theta)/(1-sintheta/costheta)#

Multi[lying numerator and denominator by #costheta#, we get

#costheta((1−cos^2theta))/(costheta-sintheta)# or

#costheta((1−cos^2theta))/(costheta-sqrt(1−cos^2theta))#