For [tex]\pi<\theta<\dfrac{3\pi}2[/tex], we expect [tex]\cos\theta<0[/tex]. Then recalling that [tex]\sin^2\theta+\cos^2\theta=1[/tex], we have
[tex]\cos\theta=-\sqrt{1-\sin^2\theta}=-\dfrac12[/tex]
Then by definition of tangent,
[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\frac{\sqrt3}2}{-\frac12}=\sqrt3[/tex]
Answer:
Option 1 -
[tex]\cos\theta=-\frac{1}{2}[/tex]
[tex]\tan\theta=\sqrt3[/tex]
Step-by-step explanation:
Given : [tex]\sin\theta=-\frac{\sqrt3}{2}[/tex] and [tex]\pi < \theta < \frac{3\pi}{2}[/tex]
To find : What are the values of [tex]\cos \theta[/tex] and [tex]\tan \theta[/tex]?
Solution :
[tex]\sin\theta=-\frac{\sqrt3}{2}[/tex]
We know, [tex]\cos\theta=\sqrt{1-sin^2\theta}[/tex]
Substitute the value of [tex]\sin\theta[/tex],
[tex]\cos\theta=\sqrt{1-(-\frac{\sqrt3}{2})^2}[/tex]
[tex]\cos\theta=\sqrt{1-\frac{3}{4}}[/tex]
[tex]\cos\theta=\sqrt{\frac{4-3}{4}}[/tex]
[tex]\cos\theta=\sqrt{\frac{1}{4}}[/tex]
[tex]\cos\theta=\frac{1}{2}[/tex]
Since, [tex]\pi < \theta < \frac{3\pi}{2}[/tex] i.e. in third quadrant
We know, [tex]\cos \theta[/tex] in third quadrant is negative.
So, [tex]\cos\theta=-\frac{1}{2}[/tex]
Now, [tex]\tan\theta=\frac{\sin\theta}{\cos\theta}[/tex]
Substitute the value of [tex]\sin\theta[/tex] and [tex]\cos\theta[/tex],
[tex]\tan\theta=\frac{-\frac{\sqrt3}{2}}{-\frac{1}{2}}[/tex]
[tex]\tan\theta=\frac{\sqrt3}{2}\times \frac{2}{1}}[/tex]
[tex]\tan\theta=\sqrt3[/tex]
We know, [tex]\tan \theta[/tex] in third quadrant is positive.
So, [tex]\tan\theta=\sqrt3[/tex]
Therefore, Option 1 is correct.
