Question

If $\cos (\alpha +\beta )=4/5and\sin (\alpha –\beta )=5/13,where0\le \alpha ,\beta \le \pi /4$, then $\tan 2\alpha$ is equal to

• A. $25/16$
• B. $56/33$
• C. $19/12$
• D. $20/7$

Answer 1

The correct option is B

$56/33$

The explanation for the correct option:

Step 1: Finding the quadrant for $(\alpha -\beta )$
Given,

$$\begin{gathered} cos(\alpha +\beta )=4/5 \\ \alpha +\beta \in Istquadrant \\ sin(\alpha –\beta )=5/13 \\ \alpha -\beta \in Istquadrant\{since0\le \alpha ,\beta \le \pi /4\} \end{gathered}$$

Step 2: Finding the value of $tan(\alpha -\beta )$

From the above,

$\begin{array}{rcl}\sin (\alpha +\beta )& =& 3/5\\ \cos (\alpha –\beta )& =& 12/13\\ Thatmeans,\tan (\alpha +\beta )& =& 3/4\\ \tan (\alpha –\beta )& =& 5/12\\ \because 2\alpha & =& (\alpha +\beta )+(\alpha –\beta )\end{array}$

Step 3: Taking tan on both sides,

$\begin{array}{rcl}\tan 2\alpha & =& \tan \left[\right(\alpha +\beta )+(\alpha –\beta \left)\right]\\ & =& \frac{\tan (\alpha +\beta )+\tan (\alpha –\beta }{1–\tan (\alpha +\beta )\tan (\alpha –\beta )}\\ & =& \frac{(3/4)+(5/12)}{1–(3/4)(5/12)}\\ & =& \frac{(9+5)/12}{(16–5)/16}\\ & =& (14/12)\times (16/11)\\ & =& 56/33\end{array}$

Hence, the correct answer is option B