Use the half angle identity
tan^2(t) = (1 - cos(2t)) / (1 + cos(2t))
Therefore,
tan^2(105) = (1 - cos(2*105)) / (1 + cos(2*105))
tan^2(105) = (1 - cos(210)) / (1 + cos(210))
Fortunately for us, we know the cosine of 210. 210 is in the third quadrant, and in radians is 7pi/6. The cosine of 7pi/6 is -sqrt(3)/2. With that said, we get
tan^2(105) = (1 - (-sqrt(3)/2) ) / (1 + (-sqrt(3)/2) )
Let's multiply top and bottom by 2, to get rid of the awkward fractions-within-fractions.
tan^2(105) = (2 - (-sqrt(3)) ) / (2 + (-sqrt(3)) )
Simplifying some more,
tan^2(105) = (2 + sqrt(3) ) / (2 - sqrt(3))
Now, we rationalize the denominator by multiplying top and bottom by 2 + sqrt(3). This effectively squares the numerator, and changes the denominator into a difference of squares.
tan^2(105) = (2 + sqrt(3))^2 / (2^2 - [sqrt(3)]^2)
tan^2(105) = (2 + sqrt(3))^2 / (4 - 3)
tan^2(105) = (2 + sqrt(3))^2 / 1
tan^2(105) = (2 + sqrt(3))^2
Which squares as
tan^2(105) = 4 + 4sqrt(3) + 3
tan^2(105) = 7 + 4sqrt(3)
Normally, when taking the square root of both sides, we end up with a positive and negative answer, in that if we were to do this to the equation above, we get
tan(105) = +/- sqrt( 7 + 4sqrt(3) )
However, we only accept one of the positive or negative answers. Here is where we ask ourselves: at 105 degrees, what is tangent? 105 degrees lies in quadrant 2, and in quadrant 2, tangent is negative. Therefore, we take the negative answer, and
tan(105) = -sqrt( 7 + 4sqrt(3) )
