So I know that $$ \tan(A+B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A) \tan(B)}, $$ but I don't know how to find $\tan(B)$ for the following problem:
If $\tan A = 2/3$ and $\sin B = 5/\sqrt{41}$ and angles $A$ and $B$ are in Quadrant I, find the value of $\tan(A+B)$.
Thanks in advance for any help.
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$\begingroup$Note that $\cos^2 B = 1 -\sin^2 B$, so you can find $\cos B$. Armed with this and the information in your question, you can find $\tan B$, and finally $\tan(A + B)$ with your identity.
$\endgroup$ 6 $\begingroup$$$\cos^2 B = 1-\sin^2B = 1-\frac{25}{41} = \frac{16}{41}$$ $$\implies \cos B = \frac{4}{\sqrt{41}}$$ $$\implies \tan B = \frac54$$ $$\tan(A+B) = \frac{\frac23+\frac54}{1-\frac23\cdot\frac54} = \boxed{\frac{23}{2}}$$
$\endgroup$ $\begingroup$First, draw out the right triangle including angle B. Then, use the Pythagorean theorem to find the length of the adjacent side. Then, use the formula tangent=(opposite/adjacent) to find tan(B) and substitute that into the formula.
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