Question 1 Report
Given that \(\tan 2A = \frac{2 \tan A}{1 - \tan^{2} A}\), evaluate \(\tan 15°\), leaving your answer in surd form.
Let \(A = 15^{o}\) so that \(2A = 30^{o}\) and \(\tan30^{o} = \dfrac{1}{\sqrt3}\). Put \(t = \tan15^{o}\).
Since \(15^{o}\) is acute, \(\tan15^{o} > 0\), so we take the positive value:
Answer Details
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