To evaluate \( \tan(105^\circ) \), we can use the angle subtraction formula for tangent:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]
We can express \( 105^\circ \) as \( 60^\circ + 45^\circ \). So we have:
- \( A = 60^\circ \) and \( B = 45^\circ \)
Now we can calculate:
\[ \tan(60^\circ) = \sqrt{3} \] \[ \tan(45^\circ) = 1 \]
Now plugging these values into the angle addition formula:
\[ \tan(105^\circ) = \tan(60^\circ + 45^\circ) = \frac{\tan(60^\circ) + \tan(45^\circ)}{1 - \tan(60^\circ) \tan(45^\circ)} = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \]
Now, to simplify \( \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \), multiply the numerator and denominator by the conjugate of the denominator, which is \( 1 + \sqrt{3} \):
\[ \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{1 - 3} = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{-2} \]
Now, simplify the numerator:
\[ (\sqrt{3} + 1)(1 + \sqrt{3}) = \sqrt{3} + 3 + 1 + \sqrt{3} = 2\sqrt{3} + 4 \]
Thus, we can write:
\[ \tan(105^\circ) = \frac{2\sqrt{3} + 4}{-2} = -\left(\sqrt{3} + 2\right) \]
So, we can express this as:
\[ \tan(105^\circ) = -2 - \sqrt{3} \]
The correct choice from the provided options, which represents \( - (2 + \sqrt{3}) \), is:
negative (2 plus start root 3 end root)
1 answer