If tan A = 3/4 and cos B = 9/41, Find tan(A+B)

Question

If \[ \tan A=\frac{3}{4} \] and \[ \cos B=\frac{9}{41} \] where \[ \pi < A < \frac{3\pi}{2} \] and \[ 0 < B < \frac{\pi}{2} \] find:

\[ \tan(A+B) \]

Given: \[ \tan A=\frac{3}{4} \]

Since \[ \pi < A < \frac{3\pi}{2} \] A lies in the third quadrant, where tangent is positive.

Using the Pythagorean triple:

\[ \sin A=-\frac{3}{5}, \qquad \cos A=-\frac{4}{5} \]

Also, \[ \cos B=\frac{9}{41} \]

Using \[ \sin^2 B+\cos^2 B=1 \]

\[ \sin B=\sqrt{1-\left(\frac{9}{41}\right)^2} \]

\[ =\sqrt{1-\frac{81}{1681}} \]

\[ =\sqrt{\frac{1600}{1681}} \]

\[ \sin B=\frac{40}{41} \]

Since \[ 0 < B < \frac{\pi}{2} \] B lies in the first quadrant, so sine is positive.

Now,

\[ \tan B=\frac{\sin B}{\cos B} \]

\[ =\frac{40/41}{9/41} \]

\[ =\frac{40}{9} \]

Using formula:

\[ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \]

\[ =\frac{\frac{3}{4}+\frac{40}{9}}{1-\left(\frac{3}{4}\times\frac{40}{9}\right)} \]

\[ =\frac{\frac{27+160}{36}}{1-\frac{120}{36}} \]

\[ =\frac{\frac{187}{36}}{\frac{36-120}{36}} \]

\[ =\frac{187}{-84} \]

Therefore,

\[ \boxed{\tan(A+B)=-\frac{187}{84}} \]

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