How do you verify the identity #cscx(cscx-sinx)+(sinx-cosx)/sinx+cotx=csc^2x#?

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Answer 1 · 2017-01-17T20:08:22

see below

Left Hand Side:

#cscx(cscx-sinx)+(sinx-cosx)/sinx + cotx = csc^2x-cscxsinx+(sinx-cosx)/sinx + cotx#

#=csc^2x-(1/sinx)sinx+(sinx-cosx)/sinx + cosx/sinx#

#=csc^2x-(1/cancel (sinx))cancel (sinx)+(sinx-cancelcosx+ cancelcosx)/sinx#

#=csc^2x-1+sinx/sinx#

Use identity: #1+cot^2x=csc^2x#

#=cot^2x+cancelsinx/cancelsinx#

#=cot^2x+1#

#=csc^2x#

#:.=# Right Hand Side