Answer:
The values of cos Θ and tan Θ are \(-\frac{1}{2}\) and √3 respectively.
Step-by-step explanation:
Given,
\(sin\theta = -\frac{\sqrt{3} }{2}\)
Since, by the trigonometric identity,
\(sin^2 \theta + cos^2\theta = 1\)
\(\implies cos^2 \theta = 1 - sin^2 \theta\)
\(\implies cos \theta = \pm \sqrt{1 - sin^2 \theta}\)
We have,
\(\pi < \theta < \frac{3\pi}{2}\)
⇒ \(\theta\) lies in third quadrant,
⇒ \(cos\theta\) is negative.
\(\implies cos \theta = - \sqrt{1 - sin^2 \theta}=-\sqrt{1 -(-\frac{\sqrt{3} }{2})^2}=-\sqrt{1 -\frac{3}{4}}=-\sqrt{\frac{1}{4}}=-\frac{1}{2}\)
Now,
\(tan \theta = \frac{sin \theta}{cos \theta}=\frac{-\sqrt{3}/2}{-1/2}=\sqrt{3}\)
