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

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

$\tan ^2(\theta)=\frac{143}{121}-1$

$\tan ^2(\theta)=\frac{22}{121}$

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