How do you solve #\frac { \tan \theta - \tan 27^ { \circ } } { 1+ \tan \theta \tan 27^ { \circ } } = 1#?

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Answer 1 · 2017-11-13T14:48:39

#\theta=72^circ+180^circ or \theta=(2pi)/5+n\pi# where #n inZZ#

#(tan\theta-tan27^circ)/(1+tan\theta*tan27^circ)=1#

Using the trig identity: #tan(A-B)=(tanA-tanB)/(1+tanAtanB)#, we can get:
#(tan\theta-tan27^circ)/(1+tan\theta*tan27^circ) = tan(\theta-27^circ)=1#

#arctan(tan(\theta-27^circ))=\theta-27^\circ#
#arctan(1)=45^circ#

#\theta-27^circ=45^circ#

#\theta=45^circ+27^circ=72^circ#

#\theta=72^circ+n180^circ or \theta=(2pi)/5+npi# where #n inZZ#

Answer 2 · 2017-11-13T14:54:53

#\theta=72^\circ#

#(tan\theta-tan(27))/(1+tan\thetatan(27))=1#

#((tan\theta-tan(27))cancel((1+tan\thetatan(27))))/cancel(1+tan\thetatan(27))=1*(1+tan\thetatan(27))#

#tan\theta-tan(27)=1+tan\thetatan(27)#

#tan\theta-tan\thetatan(27)=1+tan(27)#

#tan\theta(1-tan(27))=(1+tan(27))#

#tan\theta=(1+tan(27))/(1-tan(27))#

#\theta=arctan((1+tan(27))/(1-tan(27)))~~arctan(3.08)~~72^\circ#

Answer 3 · 2017-11-14T13:18:15

#theta=72^circ#

#(tan\theta-tan27^circ)/(1+tan\theta*tan27^circ)=1#

#tan\theta-tan27^circ=1+tan\theta*tan27^circ#

#tan\theta-tan\theta*tan27^circ=1+tan27^circ#

#tan\theta*(1-tan27^circ)=1+tan27^circ#

#tan\theta=(1+tan27^circ)/(1-tan27^circ)#

=#(cos27^circ+sin27^circ)/(cos27^circ-sin27^circ)#

=#(sin63^circ+sin27^circ)/(cos27^circ-cos63^circ)#

=#(2sin45^circ*cos18^circ)/(-2sin45^circ*sin(-18)^circ)#

=#cos18^circ/-(-sin18^circ)#

=#cot18^circ#

=#tan72^circ#

Hence #theta=72^circ#