How do you find tan 75 using the angle sum / difference identities ? I get stuck when I multiply by the conjugate (3+#sqrt3 )

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Answer 1 · 2018-04-18T23:25:52

#tan75^o=(12+6sqrt3)/6=2+sqrt3#

From the unit circle, we know the values of #tan45^o, tan30^o#:

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So, as #45^o + 30^o=75^o#, we'll use these two angles when applying the sum identity.

Recall that

#tan(a+b)=(tana+tanb)/(1-tanatanb)#

We're taking #a=45^o,b=30^o,# so

#tan(45^o + 30^o)=(tan45^o + tan30^o)/(1-tan45^otan30^o)#

#tan45^o=(sqrt2/2)/(sqrt2/2)=1#

#tan30^o=(1/2)/(sqrt3/2)=1/sqrt3=sqrt3/3#

Thus,

#tan75^o=(1+sqrt3/3)/(1-sqrt3/3)#

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

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

Multiply by the conjugate of #3-sqrt3,# which is #3+sqrt3:#

#=(3+sqrt3)/(3-sqrt3)*(3+sqrt3)/(3+sqrt3)#

#(3+sqrt3)(3+sqrt3)=9+6sqrt3+(sqrt3)^2=9+6sqrt3+3=12+6sqrt3#

#(3+sqrt3)(3-sqrt3)=9-(sqrt3)^2=9-3=6#

Thus,

#tan75^o=(12+6sqrt3)/6=2+sqrt3#