Question

Let 0 be an angle in quadrant II such that cos 0

Answer 1

we have that

cos(theta)=-5/6

step 1

Find out sin(theta)

Remember that

$\sin ^2(\theta)+\cos ^2(\theta)=1$

substitute given value

$\sin ^2(\theta)+(-\frac{5}{6})^2=1$$\sin ^2(\theta)^{}=1-\frac{25}{36}$$\sin ^{}(\theta)^{}=\frac{\sqrt[]{11}}{6}$

The value of sin(theta) is positive because the angle theta lies on the II quadrant

step 2

Find out csc(theta)

$\csc (\theta)=\frac{1}{\sin (\theta)}$$\csc (\theta)=\frac{6}{\sqrt[]{11}}$

simplify

$\csc (\theta)=\frac{6\sqrt[]{11}}{11}$

step 3

Find out tan(\theta)

$\tan (\theta)=\frac{\sin (\theta)}{\cos (\theta)}$$\tan (\theta)=-\frac{\sqrt[]{11}}{5}$