Question

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

Answer 1

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

Explanation:

`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)`