Q.5 (ii) \(\alpha\) is acute angle. If \(3\sin\alpha - 4\cos\alpha = 0\), then find the values of \(\tan\alpha\) and \(\sec\alpha\).
Best answer
Answer: \(\tan\alpha = \dfrac{4}{3}\) and \(\sec\alpha = \dfrac{5}{3}\)
\(3\sin\alpha - 4\cos\alpha = 0\), so \(3\sin\alpha = 4\cos\alpha\).
Dividing both sides by \(3\cos\alpha\) (as \(\alpha\) is acute, \(\cos\alpha \ne 0\)):
\(\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{4}{3}\).
Now \(1 + \tan^2\alpha = \sec^2\alpha\):
\(\sec^2\alpha = 1 + \dfrac{16}{9} = \dfrac{25}{9}\).
\(\alpha\) is acute, so \(\sec\alpha\) is positive: \(\sec\alpha = \dfrac{5}{3}\).
(Check: with \(\tan\alpha = \dfrac{4}{3}\), \(\sin\alpha = \dfrac{4}{5}\) and \(\cos\alpha = \dfrac{3}{5}\), and \(3 \times \dfrac{4}{5} - 4 \times \dfrac{3}{5} = 0\).)
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