Question

If \[\sin A = \frac{1}{2}, \cos B = \frac{\sqrt{3}}{2}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
tan (A + B)

Answer 1

\[\text{ Given: }\sin A = \frac{1}{2}\text{ and }\cos B = \frac{\sqrt{3}}{2}\]
\[\text{ Here,} \frac{\pi}{2} < A < \pi\text{ and }0 < B < \frac{\pi}{2} . \]
That is, A is in thesecond quadrant and B is in the first quadrant .
We know that in the second quadrant, sine function is positive and cosine and tan functions are negative
In the first quadrant, all T - functions are positive . 
Therefore,
\[\cos A = - \sqrt{1 - \sin^2 A} = - \sqrt{1 - \left( \frac{1}{2} \right)^2} = - \sqrt{1 - \frac{1}{4}} = - \sqrt{\frac{3}{4}} = \frac{- \sqrt{3}}{2}\]
\[\tan A = \frac{\sin A}{\cos A} = \frac{\frac{1}{2}}{\frac{- \sqrt{3}}{2}} = \frac{- 1}{\sqrt{3}}\]
\[\sin B = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left( \frac{\sqrt{3}}{2} \right)^2} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\]
\[\tan B = \frac{\sin B}{\cos B} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}\]
Now, 
\[ \tan\left( A + B \right) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\]
\[ = \frac{\frac{- 1}{\sqrt{3}} + \frac{1}{\sqrt{3}}}{1 - \frac{- 1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}}\]
\[ = \frac{0}{1 + \frac{1}{3}} = 0\]