How do you verify the identity #[(cosx cotx)/1-sin x]-1=cscx#?

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Answer 1 · 2015-06-12T17:09:19

Is there a typo here? These are not equal.

With #cotx = cosx/sinx# and #cscx = 1/sinx#:

#cosxcotx - sinx - 1 = cscx#

#cos^2x/sinx - sinx - 1 = cscx#

#cos^2x/sinx - sinx - 1 = 1/sinx#

#cos^2x - sin^2x - sinx = 1#

#cos^2x - sin^2x - sinx = sin^2x + cos^2x#

#cos^2x ne [sinx + 2sin^2x] + cos^2x#