Question

Let theta be an angle in quadrant II such that cos theta = -2/3

Answer 1

Answer:

$tan\theta{3}=-\frac{\sqrt5}{2}$

$cosec\theta=\frac{3}{\sqrt5}$

Step-by-step explanation:

We are given that $\theta$ be an  angle in quadrant II and $cos\theta=-\frac{2}{3}$

We have to find the exact values of $cosec\theta$ and $tan\theta$.

$sec\theta=\frac{1}{cos\theta}$

Then substitute the value of cos theta and we get

$sec\theta=\frac{1}{-\frac{2}{3}}$

$sec\theta=-\frac{3}{2}$

Now, $1+tan^2\theta=sec^2\theta$

$tan^2\theta=sec^2\theta-1$

Substitute the value of sec theta then we get

$tan^2\theta= (-\frac{3}{2})^2-1$

$tan^2\theta=\frac{9}{4}-1=\frac{9-4}{4}=\frac{5}{4}$

$tan\theta=\sqrt{\frac{5}{4}}=-\frac{\sqrt5}{2}$

Because$tan\theta$ in quadrant II is negative.

$sin^2\theta=1-cos^2\theta$

$sin^2\theta=1-(\farc{-2}{3})^2$

$sin^2\theta=1-\frac{4}{9}$

$sin^2\theta=\frac{9-4}{9}=\frac{5}{9}$

$sin\theta=\sqrt{\frac{5}{9}}$

$sin\theta=\frac{\sqrt5}{3}$

Because in quadrant II $sin\theta$ is positive.

$cosec\theta=\frac{1}{sin\theta}=\frac{1}{\frac{\sqrt5}{3}}$

$cosec\theta=\frac{3}{\sqrt5}$

$cosec\theta$ is positive in II quadrant.

Answer 2

Answer:

So we have $\csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}$.

Step-by-step explanation:

Ok so we are in quadrant 2, that means sine is positive while cosine is negative.

We are given $\cos(\theta)=\frac{-2}{3}(\frac{\text{adjacent}}{\text{hypotenuse}})$.

So to find the opposite we will just use the Pythagorean Theorem.

$a^2+b^2=c^2$

$(2)^2+b^2=(3)^2$

$4+b^2=9$

$b^2=5$

$b=\sqrt{5}$  This is the opposite side.

Now to find $\csc(\theta)$ and $\tan(\theta)$.

$\csc(\theta)=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{3}{\sqrt{5}}$.

Some teachers do not like the radical on bottom so we will rationalize the denominator by multiplying the numerator and denominator by sqrt(5).

So $\csc(\theta)=\frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{3 \sqrt{5}}{5}$.

And now $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt{5}}{-2}=\frac{-\sqrt{5}}{2}$.

So we have $\csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}$.