Hello, Keith!
Your arithmetic is a bit off . . .
Given: $\displaystyle \sin\theta = \frac{3}{4},\;\theta $ in Q2. . Find $\displaystyle \tan\theta$
Using your approach, I'd do it like this . . .
. . $\displaystyle \cos^2\theta \:=\:1  \sin^2\theta \;= \; 1  \left(\frac{3}{4}\right)^2 \:=\:1  \frac{9}{16} \:=\:\frac{7}{16}$
Then: .$\displaystyle \cos\theta \:=\:\pm\frac{7}{4} $
Since $\displaystyle \theta$ is in Q2, cosine is negative: .$\displaystyle \cos\theta = \frac{\sqrt{7}}{4}$
Therefore: .$\displaystyle \tan\theta \:=\:\frac{\sin\theta}{\cos\theta} \:=\:\frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} \:=\:\boxed{\frac{3}{\sqrt{7}}} $
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I prefer to do it like this . . .
We're given: .$\displaystyle \sin\theta = \frac{3}{4},\;\theta\text{ in Q2.}$
Since $\displaystyle \sin\theta = \frac{opp}{hyp}$, I can make a sketch. Code:
* 
: \ 
3 : \4 
: \ θ
+*

We have: .$\displaystyle opp = 3,\;hyp = 4.$
Using Pythagorus, we find that: $\displaystyle adj = \sqrt{7}$
. . And now we can write any of the trig functions.