To find the exact value of [tex]\( \tan 105^\circ \)[/tex], we can use the tangent addition formula. Recall that we can find [tex]\( \tan(105^\circ) \)[/tex] by expressing 105° as the sum of two angles whose tangents we know. One convenient way to do this is:
[tex]\[ 105^\circ = 60^\circ + 45^\circ \][/tex]
Using the tangent addition formula:
[tex]\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \][/tex]
Let [tex]\( A = 60^\circ \)[/tex] and [tex]\( B = 45^\circ \)[/tex]. We know that:
[tex]\[ \tan 60^\circ = \sqrt{3} \][/tex]
[tex]\[ \tan 45^\circ = 1 \][/tex]
Substituting these values into the formula, we get:
[tex]\[ \tan(105^\circ) = \frac{\tan 60^\circ + \tan 45^\circ}{1 - \tan 60^\circ \cdot \tan 45^\circ} \][/tex]
[tex]\[ \tan(105^\circ) = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} \][/tex]
[tex]\[ \tan(105^\circ) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \][/tex]
To simplify this, we rationalize the denominator. Multiplying the numerator and denominator by the conjugate of the denominator, [tex]\( 1 + \sqrt{3} \)[/tex]:
[tex]\[ \tan(105^\circ) = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \][/tex]
First, expand the numerator:
[tex]\[ (\sqrt{3} + 1)(1 + \sqrt{3}) = \sqrt{3} \cdot 1 + \sqrt{3} \cdot \sqrt{3} + 1 \cdot 1 + 1 \cdot \sqrt{3} \][/tex]
[tex]\[ = \sqrt{3} + 3 + 1 + \sqrt{3} \][/tex]
[tex]\[ = 4 + 2\sqrt{3} \][/tex]
Expand the denominator:
[tex]\[ (1 - \sqrt{3})(1 + \sqrt{3}) = 1^2 - (\sqrt{3})^2 \][/tex]
[tex]\[ = 1 - 3 \][/tex]
[tex]\[ = -2 \][/tex]
Now, combining the numerator and the denominator:
[tex]\[ \tan(105^\circ) = \frac{4 + 2\sqrt{3}}{-2} \][/tex]
We can further simplify this:
[tex]\[ \tan(105^\circ) = -2 - \sqrt{3} \][/tex]
So the exact value of [tex]\( \tan 105^\circ \)[/tex] is:
[tex]\[ \boxed{-2 - \sqrt{3}} \][/tex]
