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

1 Answer
Jan 17, 2017

see below

Explanation:

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