How do you express #tan theta - cot theta # in terms of #cos theta #?

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Answer 1 · 2016-03-25T02:32:06

We must use the quotient identities, #tantheta = sintheta/costheta# and #cottheta = costheta/sintheta#

= #sintheta/costheta - costheta/sintheta#

=#sin^2theta/(costhetasintheta) - cos^2theta/(costhetasintheta)#

Use the pythagorean identity #sin^2theta + cos^2theta = 1# to simplify further.

#= (1 - cos^2theta - cos^2theta)/(costhetasintheta)#

#= (1 - 2cos^2theta)/(costhetasintheta)#

This is simplest form; we can't get rid of the sin (pardon the unintended pun!).

Hopefully this helps!

Answer 2 · 2016-03-28T13:24:35

#sqrt(1-cos^2theta)/costheta-costheta/sqrt(1-cos^2theta)#

Express first in terms of #sintheta# and #costheta#.

#=sintheta/costheta-costheta/sintheta#

Now, to turn the #sintheta# terms into #costheta# terms, use the Pythagorean identity:

#sin^2theta+cos^2theta=1#

#sin^2theta=1-cos^2theta#

#sintheta=sqrt(1-cos^2theta)#

Substitute this into the expression we originally made:

#=sqrt(1-cos^2theta)/costheta-costheta/sqrt(1-cos^2theta)#