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Question:

If $\tan A=\frac{3}{4}, \cos B=\frac{9}{41}$, where $\pi<A<\frac{3 \pi}{2}$ and $0<B<\frac{\pi}{2}$, find $\tan (A+B)$.

Solution:

Given:

$\tan A=\frac{3}{4}$ and $\cos B=\frac{9}{41}$

Here, $\pi<A<\frac{3 \pi}{2}$ and $0<B<\frac{\pi}{2}$.

That is, $A$ is in third quadrant and $B$ is in first qudrant.

We know that tan function is positive in first and third quadrant $s$,

and in the first quadrant, $\sin e$ function is also positive.

Therefore, $\sin B=\sqrt{1-\cos ^{2} B}$

$=\sqrt{1-\left(\frac{9}{41}\right)^{2}}$

$=\sqrt{1-\frac{81}{1681}}$

$=\sqrt{\frac{1600}{1681}}$

$=\frac{40}{41}$

And $\tan B=\frac{\sin B}{\cos B}$

$=\frac{{ }^{40} /{ }_{41}}{9 / 41}=\frac{40}{9}$

Therefore, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$

$=\frac{\frac{3}{4}+\frac{40}{9}}{1-\frac{3}{4} \times \frac{40}{9}}$

$=\frac{\frac{187}{36}}{\frac{-84}{36}}$

$=\frac{-187}{84}$

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