Question

If $\cos (\alpha +\beta )=\frac{4}{5}$, $\sin (\alpha -\beta )=\frac{5}{13}$ and $\alpha ,\beta$ lie between $0$ and $\frac{\pi }{4}$, then $\tan 2\alpha =?$

• A. $\frac{16}{63}$
• B. $\frac{56}{33}$
• C. $\frac{28}{33}$
• D. None of these

Answer 1

The correct option is B

$\frac{56}{33}$

Explanation for the correct option:

Step 1. Find the value of $\tan 2\alpha$:

Given, $\cos (\alpha +\beta )=\frac{4}{5}$

$\Rightarrow$ $\sin (\alpha +\beta )=\frac{3}{5}$

$\sin (\alpha -\beta )=\frac{5}{13}$

$\Rightarrow$ $\cos (\alpha -\beta )=\frac{12}{13}$

Now, we can write

$2\alpha =\alpha +\beta +\alpha –\beta$

Step 2. Take $"\tan "$ on both sides, we get

$\tan 2\alpha =\tan (\alpha +\beta +\alpha –\beta )$

$\tan 2\alpha =\frac{\left[\tan \right(\alpha +\beta )+\tan (\alpha –\beta \left)\right]}{[1–\tan (\alpha +\beta )\tan (\alpha –\beta \left)\right]}$ …(1) $\left[\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\left(\theta +\varphi \right)\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\varphi }}{\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\theta }\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\varphi }}\right]$

Also,

$\begin{array}{rcl}\tan (\alpha +\beta )& =& \frac{\sin (\alpha +\beta )}{\cos (\alpha +\beta )}\\ & =& \frac{3/5}{4/5}\\ & =& \frac{3}{4}\end{array}$

$\begin{array}{rcl}\tan (\alpha –\beta )& =& \frac{\sin (\alpha –\beta )}{\cos (\alpha –\beta )}\\ & =& \frac{5/13}{12/13}\\ & =& \frac{5}{12}\end{array}$

Step 3. Put these values in equation (1), we get

$\begin{array}{rcl}\therefore \tan 2\alpha & =& \frac{(3/4)+(5/12)}{1–(3/4)(5/12)}\\ & =& \frac{(9+5)/12}{(48–15)/48}\\ & =& \frac{\mathbf{56}}{\mathbf{33}}\end{array}$

Hence, Option ‘B’ is Correct.