How do you verify the identity #(costhetacottheta)/(1-sintheta)-1=csctheta#?

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Answer 1 · 2017-01-17T07:33:52

See proof below

We need

#cottheta=costheta/sintheta#

#csctheta=1/sintheta#

#cos^2theta+sin^2theta=1#

#a^2-b^2=(a+b)(a-b)#

Therefore,

#LHS=(costhetacottheta)/(1-sintheta)-1#

#=(costheta*costheta/sintheta)/(1-sintheta)-1#

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

#=(1-sin^2theta)/(sintheta(1-sintheta))-1#

#=((1+sintheta)cancel(1-sintheta))/(sinthetacancel(1-sintheta))-1#

#=(1+sintheta)/sintheta-1#

#=(1+sintheta-sintheta)/sintheta#

#=1/sintheta#

#=csctheta#

#=RHS#

#QED#