Discussion
\(tan15^{\circ} + tan75^{\circ}\)=
Answer: D
\(\tan 15^{\circ} + \tan 75^{\circ} = \) 4.
Since \(\tan 75^{\circ} = \cot 15^{\circ}\), write \(\theta = 15^{\circ}\):
\[ \begin{aligned} \tan\theta + \cot\theta &= \frac{\sin^{2}\theta + \cos^{2}\theta}{\sin\theta\cos\theta} \\ &= \frac{2}{\sin 2\theta} = \frac{2}{\sin 30^{\circ}} = 4 \end{aligned} \]
Directly: \(\tan 15^{\circ} = 2 - \sqrt{3}\) and \(\tan 75^{\circ} = 2 + \sqrt{3}\), which add to 4. Their product is 1, not their sum; 2 is what you get by taking each tangent as 1, and 0 would need the two angles to be negatives of each other.
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