How do you find the exact value of #tan((7pi)/12)#?

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Answer 1 · 2016-05-22T13:15:01

#tan((7pi)/12)=-(2+sqrt3)#

#tan((7pi)/12)=tan(pi-(5pi)/12)#

= #-tan((5pi)/12)

= #-tan((3pi)/12+(2pi)/12)#

= #-tan(pi/4+pi/6)#

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

= #-(tan(pi/4)+tan(pi/6))/(1-tan(pi/4)tan(pi/6)#

= #-(1+1/sqrt3)/(1-1xx1/sqrt3)#

Multiplying numerator and denominator by #sqrt3#

= #-(sqrt3+1)/(sqrt3-1)=-(sqrt3+1)^2/((sqrt3-1)(sqrt3+1))#

= #-(3+1+2sqrt3)/(3-1)=-(2+sqrt3)#