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)+(-(5)/(6))^2=1`
`\sin ^2(\theta)^{}=1-(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)=(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)=(\sin (\theta))/(\cos (\theta))`
`\tan (\theta)=-\frac{\sqrt[]{11}}{5}`