Given that \(\tan 2A = \frac{2 \tan A}{1 - \tan^{2} A}\), evaluate \(\tan 15°\), leaving your answer in surd form.
Question 1 Report
Given that \(\tan 2A = \frac{2 \tan A}{1 - \tan^{2} A}\), evaluate \(\tan 15°\), leaving your answer in surd form.
Answer Details
Let \(A = 15^{o}\) so that \(2A = 30^{o}\) and \(\tan30^{o} = \dfrac{1}{\sqrt3}\). Put \(t = \tan15^{o}\).
\[\tan30^{o} = \frac{2\tan15^{o}}{1 - \tan^2 15^{o}} \;\Rightarrow\; \frac{1}{\sqrt3} = \frac{2t}{1 - t^2}\] \[1 - t^2 = 2\sqrt3\,t \;\Rightarrow\; t^2 + 2\sqrt3\,t - 1 = 0\] \[t = \frac{-2\sqrt3 \pm \sqrt{12 + 4}}{2} = \frac{-2\sqrt3 \pm 4}{2} = -\sqrt3 \pm 2\]Since \(15^{o}\) is acute, \(\tan15^{o} > 0\), so we take the positive value:
\[\tan15^{o} = 2 - \sqrt3\]Download The App On Google Playstore
Everything you need to excel in your exams
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning