How do you verify #cscx-1=cot^2x/(cscx+1)#?

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Answer 1 · 2016-11-12T14:47:43

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

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

#(1 - sinx)/sinx= cos^2x/sin^2x xx sinx/(1 + sinx)#

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

#(1 - sinx)/sinx = ((1 + sinx)(1 - sinx))/sinx xx 1/(1 + sinx)#

#(1- sinx)/sinx = (1 - sinx)/sinx#

#LHS = RHS#

Identity Proved!

Hopefully this helps!