Question

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

 

Answer 1

Given:
$$\tan A=\frac{3}{4}\text{ and }\cos B=\frac{9}{41}$$
$$\text{ Here,}\pi <A<\frac{3\pi }{2}\text{ and }0<B<\frac{\pi }{2}.$$
That is, A is in third quadrant and B is in first qudrant . 
We know that tan function is positive in first and third quadrants, and in the first quadrant, \sine function is also positive . 
$$\text{ Therefore, }\sin B=\sqrt{1-{\cos }^{2}B}$$
$$=\sqrt{1-{\left(\frac{9}{41}\right)}^2}$$
$$=\sqrt{1-\frac{81}{1681}}$$
$$=\sqrt{\frac{1600}{1681}}$$
$$=\frac{40}{41}$$
$$\text{ And }\tan B=\frac{\sin B}{\cos B}$$
$$=\frac{\frac{40}{41}}{\frac{9}{41}}=\frac{40}{9}$$
$$\text{Therefore, }\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
$$=\frac{\frac{3}{4}+\frac{40}{9}}{1-\frac{3}{4}\times \frac{40}{9}}$$
$$=\frac{\frac{187}{36}}{\frac{-84}{36}}$$
$$=\frac{-187}{84}$$