Question

If \[\sin A = \frac{1}{2}, \cos B = \frac{12}{13}\], where \[\frac{\pi}{2}\]< A < π and \[\frac{3\pi}{2}\] < B < 2π, find tan (A − B).

Answer 1

Given:
\[\sin A = \frac{1}{2}\text{ and }\cos B = \frac{12}{13}\]
\[\text{ Here, }\frac{\pi}{2} < A < \pi \text{ and }\frac{3\pi}{2} < B < 2\pi . \]
That is, A is in the second quadrant and B is in the fourth quadrant . 
We know that in the second quadrant, sine function is positive and cosine and tan functions are negative . 
In the fourth quadrant, sine and tan functions are negative and cosine function is positive . 
Therefore, 
\[\cos A = - \sqrt{1 - \sin^2 A} = - \sqrt{1 - \left( \frac{1}{2} \right)^2} = - \sqrt{1 - \frac{1}{4}} = - \sqrt{\frac{3}{4}} = \frac{- \sqrt{3}}{2}\]
\[\tan A = \frac{\sin A}{\cos A} = \frac{\frac{1}{2}}{\frac{- \sqrt{3}}{2}} = \frac{- 1}{\sqrt{3}}\]
\[\sin B = - \sqrt{1 - \cos^2 B} = - \sqrt{1 - \left( \frac{12}{13} \right)^2} = - \sqrt{1 - \frac{144}{169}} = - \sqrt{\frac{25}{169}} = \frac{- 5}{13}\]
\[\tan B = \frac{\sin B}{\cos B} = \frac{- \frac{5}{13}}{\frac{12}{13}} = \frac{- 5}{12}\]
\[\text{ Now, }\tan\left( A - B \right) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\]
\[ = \frac{\frac{- 1}{\sqrt{3}} - \frac{- 5}{12}}{1 + \frac{- 1}{\sqrt{3}} \times \frac{- 5}{12}}\]
\[ = \frac{\frac{- 12 + 5\sqrt{3}}{12\sqrt{3}}}{\frac{12\sqrt{3} + 5}{12\sqrt{3}}} = \frac{5\sqrt{3} - 12}{5 + 12\sqrt{3}}\]