How do you simplify the expression #(tant)/(tant+cott)#?

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Answer 1 · 2016-10-15T14:14:37

This expression can be simplified to #sin^2t#.

For this problem, the following identities will be important:

#•tantheta = sintheta/costheta#

#•cot theta = 1/tantheta= 1/(sintheta/costheta) = costheta/sintheta#

Start simplifying:

#=(sint/cost)/(sint/cost + cost/sint)#

#= (sint/cost)/((sin^2t + cos^2t)/(costsint))#

We can now apply the identity #sin^2alpha + cos^2alpha = 1# to the numerator of the lower expression.

#=(sint/cost)/(1/(costsint))#

#=sint/cost xx (costsint)#

#= sint xx sint#

#= sin^2t#

Hopefully this helps!