Chapter: Trigonometry· Mathematics Part 2 (Geometry)
Answer: \(\tan\alpha = \dfrac{4}{3}\), \(\sec\alpha = \dfrac{5}{3}\)
\(3\sin\alpha = 4\cos\alpha\). Dividing both sides by \(3\cos\alpha\): \(\dfrac{\sin\alpha}{\cos\alpha} = \dfrac{4}{3}\) \(\therefore \tan\alpha = \dfrac{4}{3}\).
\(\sec^2\alpha = 1 + \tan^2\alpha = 1 + \dfrac{16}{9} = \dfrac{25}{9}\)
α is acute, so \(\sec\alpha\) is positive: \(\sec\alpha = \dfrac{5}{3}\).