Simplify #(tan((5pi)/14)+tan(pi/14))/(1-tan((5pi)/14)tan(pi/14))#?

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Answer 1 · 2017-03-29T15:58:41

#(tan((5pi)/14)+tan(pi/14))/(1-tan((5pi)/14)tan(pi/14))=tan((3pi)/7)#

Perhaps questioner means #(tan((5pi)/14)+tan(pi/14))/(1-tan((5pi)/14)tan(pi/14))#

For this we can use the identity

#(tanA+tanB)/(1-tanAtanB)=tan(A+B)#

Here putting #A=(5pi)/14# and #B=pi/14#, we get

#(tan((5pi)/14)+tan(pi/14))/(1-tan((5pi)/14)tan(pi/14))#

and hence

#(tan((5pi)/14)+tan(pi/14))/(1-tan((5pi)/14)tan(pi/14))=tan((5pi)/14+pi/14)#

= #tan((6pi)/14)#

= #tan((3pi)/7)#