How do you rationalize the denominator for #sqrt(cosx/tanx)#?

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Answer 1 · 2016-09-20T02:08:13

#sqrt((cosx/tanx)(1))=sqrt((cosx/tanx)(cotx/cotx))=sqrt(cosxcotx) = |cosx|sqrt(cscx)#

Starting with the original:

#sqrt(cosx/tanx)#

At this point we can't do anything about the square root, so let's focus on getting the tangent ratio out from the denominator.

Remember that #cotx=1/tanx#, so we can:

#sqrt((cosx/tanx)(1))=sqrt((cosx/tanx)(cotx/cotx))=sqrt(cosxcotx)#

We can apply identities to simplify this:

#sqrt(cosxcotx) = sqrt(cosx xx cosx/sinx) = sqrt(cos^2x/sinx) = |cosx|sqrt(1/sinx) = |cosx|sqrt(cscx)#