Suppose sin(theta) = -2/5 and cos(theta) is > 0. What is the value of tan(theta)?

[tex]\begin{gathered} \sin \theta=-\frac{2}{5} \\ \text{ Taking the }\sin ^{-1}\text{ of both sides, we get} \\ \\ \theta=\sin ^{-1}(-\frac{2}{5})=-23.5782^o \end{gathered}[/tex]

Recall:

The sine of angles is only negative in the 3rd and 4th quadrants.

And the formula for the 3rd quadrant is:

[tex]\theta-180^o,180^o<\theta\leq270^o[/tex]

While the formula for the 4th quadrant is:

[tex]360^o-\theta,270^o<\theta\leq360^o[/tex]

So,

[tex]-\sin (\theta-180)=-\sin 23.5782^o[/tex]

This implies that:

[tex]\theta-180=23.5782\text{ (having divided both sides by -}\sin )[/tex]

Collecting the like terms, we get

[tex]\begin{gathered} \theta=23.5782+180=203.5782^o \\ \text{ }\theta\ne203.5782^o\text{ (since }\cos 203.5782^o\text{ is less than zero)} \\ \cos 203.5782=-0.9165<0 \\ \text{Therefore,} \\ \theta\ne203.5782^o \end{gathered}[/tex]

We shall use the 4th quadrant formula.

[tex]-\sin (360^o-\theta)=-\sin 23.5782^o[/tex]

Dividing both sides by -sin, we get

[tex]\begin{gathered} 360^o-\theta=23.5782 \\ \text{Collecting the like terms, we get} \\ 360-23.5782=\theta \\ \text{ So,} \\ \theta=336.4218^o \end{gathered}[/tex]

To check that it satisfies the given condition that says cos(theta) is greater zero, we have:

[tex]\cos 336.4218^o=0.9165>0\text{ (Condition satisfied)}[/tex]

So, we have that:

[tex]\theta=336.4218^o[/tex]

Step 2:

We shall find the value of tan(theta):

[tex]\tan \theta=\tan (336.4218^o)=-0.4364[/tex]

Therefore, the correct answer is -0.4364