Question

Hi, can you help me to solve this exercise, please

Answer 1

In this problem

the angle theta lies on the II quadrant

so

the tangent is negative

Remember that

`1+\tan ^2(\theta)=\sec ^2(\theta)`

substitute the given value

`1+\tan ^2(\theta)=(-\frac{\sqrt[]{143}}{11})^2`

`\tan ^2(\theta)=(143)/(121)-1`

`\tan ^2(\theta)=(22)/(121)`

`\tan ^{}(\theta)=-\frac{\sqrt[]{22}}{11}`