How do you solve for tan 75 when using sum and difference formula?

Question

I know that tan 75= tan (45+30)=tan (A+B)
so, (tan 45+tan 30) / (1-tan 45 tan 30) =
(1+1/radical 3) / (1-1 times 1/radical 3) then from there my math gets all messed up. Can anyone show me the correct way to finish this and end up with the answer: 2+radical 3? Thank you!!!!!!!!!!!!!

Answer 1

tan(75)
= tan(45 + 30)
=(tan 45 + tan 30) / (1- tan 45 tan 30)
= (1 + 1/sqrt(3)) / (1 - 1/sqrt(3))

You can rationalize the denominators of both of those which gives you:

(1 + sqrt(3)/3) / (1 - sqrt(3)/(3))

Finding common denominators gives you:

[(3 + sqrt(3)) / 3] / [(3 - sqrt(3) / 3].

You can use the "keep change flip" rule for complex fractions to get:

[(3 + sqrt(3)) / 3] * [3 / (3 - sqrt(3))].

The 3's cancel and leave you with:

[3 + sqrt(3)] / [3 - sqrt(3)]

We can rationalize this again by multiplying top and bottom by 3+sqrt(3) which leaves us:

(3 + sqrt(3))^2 / [(3 + sqrt(3))(3 - sqrt(3))]
= (9 + 6sqrt(3) + 3) / [9 - 3]
= (12 + 6sqrt(3)) / (6)
2 + sqrt(3)

Does that make sense?

And note: my sqrt(3) is the same as your radical(3).

:)

Answer 2

Tan 75

Answer 3

When you have a complex fraction, multiply top and bottom by the common denominator:

[1 + 1/√3] / [1 - 1/√3] =
[1 + 1/√3](√3) / [1 - 1/√3](√3) =
[√3 + 1] / [√3 - 1] =

and here multiply top and bottom by the conjugate of the bottom:

[(√3 + 1)(√3 + 1)] / [(√3 - 1)(√3 + 1)] =
[3 + 2√3 + 1] / [ 3 - 1 ] =
[4 + 2√3 ] / 2 =
2 + √3

Answer 4

Your last step is
(1+1/√3) / (1-1/√3)
= (√3+1)/(√3-1) ......(taking LCM of numerator & denominator and simplifying)
= [(√3+1)/(√3-1)]/[(√3+1)/(√3+1)]
= (√3+1)^2/[(√3-1)(√3+1)]
= (3+1+2√3)/(3-1)
= (4+2√3)/2
= 2(2+√3)/2
= 2+√3
I hope this helps.