How do you solve \frac { \tan \theta - \tan 27^ { \circ } } { 1+ \tan \theta \tan 27^ { \circ } } = 1tanθtan271+tanθtan27=1?

3 Answers
Nov 13, 2017

\theta=72^circ+180^circ or \theta=(2pi)/5+n\piθ=72+180orθ=2π5+nπ where n inZZ

Explanation:

(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

Nov 13, 2017

\theta=72^\circ

Explanation:

(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

Nov 14, 2017

theta=72^circ

Explanation:

(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