Question

PLEASE HELP ME!! LOTS OF POINTS!!

Answer 1

Answer:

$cos\theta=-\frac{1}{2}$

$tan\theta=\sqrt3$

Step-by-step explanation:

We are given that

$sin\theta=-\frac{\sqrt3}{2}$

Where $\theta$ lies between $\pi\;and\;\frac{3\pi}{2}$

We have to find the value of $cos\theta$ and $tan\theta$

$\theta$ lies in third quadrant

$cos\theta=\sqrt{1-sin^2\theta}$

$cos\theta=\sqrt{1-(\frac{\sqrt3}{2})^2}=\sqrt{1-\frac{3}{4}}=\sqrt{\frac{1}{4}}=\pm\frac{1}{2}$

$cos\theta$ is negative in third quadrant

Therefore, $cos\theta=-\frac{1}{2}$

$tan\theta=\frac{sin\theta}{cos\theta}=\frac{\frac{-\sqrt3}{2}}{\frac{-1}{2}}$

$tan\theta=\frac{\sqrt3}{2}\times 2$

$tan\theta=\sqrt3$

$tan\theta$ is positive in third quadrant

Hence, $tan\theta=\sqrt3$

Answer 2

Answer:

Just find sin^2theta and change them with the help of known identities.

You can also use Pythagoras theorm. It will work..

Hope it helps