How do you prove #{cos(x)cot(x)}/{1-sin(x)} - 1 = csc(x)#?

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Answer 1 · 2016-05-28T05:44:58

#LHS={cos(x)cot(x)}/{1-sin(x)} - 1 #

#={cos(x)cot(x)}/{1-sin(x)} sinx/sinx- 1 #

#=[(cosx*cosx)/cancelsinx*cancelsinx}/"(1-sinx)sinx"- 1 #

#=cos^2x/"(1-sinx)sinx"- 1 #

#=(1-sin^2x)/"(1-sinx)sinx"- 1 #

#=((1+sinx)cancel((1-sinx)))/(cancel((1-sinx))*sinx)- 1 #

#=((1+sinx)-sinx)/sinx#
#=(1+cancelsinx-cancelsinx)/sinx=1/sinx=cscx=RHS#
proved