How do you verify #secx-cosx=sinx/cotx#?

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Answer 1 · 2016-04-24T16:24:41

Use the following identities:

#secx = 1/cosx#

#cotx = cosx/sinx#

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

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

Applying the pythagorean identity #1 - cos^2x = sin^2x#:

#sin^2x/cosx = sin^2x/cosx#

Thus, your identity is proved!