How do you verify the identity #cot^2t/csct=csct-sint#?

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Answer 1 · 2017-01-17T19:51:44

see below

Left Hand Side:

#cot^2 t / csc t = (cos^2 t / sin^2t)/(1/sint)#

#= cos^2 t / sin^2t * sint /1#

#= cos^2 t / sin^cancel2t * cancel (sint )/1#

#=cos^2 t /sint#

Use identity: #sin^2t+cos^2t=1#

#=(1-sin^2t)/sint#

#=1/sint - sin^2t / sint#

#=1/sint - sin^cancel 2t / cancel(sint)#

#=csct - sint#

#:.=# Right Hand Side