How do you use the sum or difference identities to find the exact value of #tan(pi/12)#?

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Answer 1 · 2016-11-16T09:55:45

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

As #pi/3-pi/4=(4pi)/12-(3pi)/12=pi/12#

#tan(pi/12)=tan(pi/3-pi/4)#

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

Hence, #tan(pi/3-pi/4)=(tan(pi/3)-tan(pi/4))/(1+tan(pi/3)tan(pi/4))#

= #(sqrt3-1)/(1+sqrt3xx1)#

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

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

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

= #2-sqrt3#