Discussion
If cosθ=−3/5 then find tanθ, where θ is an angle in the second quadrant:
Answer: A
\(\tan\theta = -\frac{4}{3}\).
From \(\cos\theta = -\frac{3}{5}\), use \(\sin^{2}\theta + \cos^{2}\theta = 1\):
\[ \begin{aligned} \sin^{2}\theta &= 1 - \frac{9}{25} = \frac{16}{25} \\ \sin\theta &= \frac{4}{5} \\ \tan\theta &= \frac{4/5}{-3/5} = -\frac{4}{3} \end{aligned} \]
Sine is positive in the second quadrant, which is why \(\sin\theta\) is \(+\frac{4}{5}\), and tangent, sine over cosine, is negative there. That rules out \(\frac{4}{3}\). \(-\frac{4}{5}\) is a value of sine with the wrong sign, and \(\frac{3}{5}\) is cosine without its sign.
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