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Question

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

A

$\frac{16}{63}$

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B

$\frac{56}{33}$

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C

$\frac{28}{33}$

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D

None of these

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Solution

## The correct option is B $\frac{56}{33}$Explanation for the correct option:Step 1. Find the value of $\mathrm{tan}2\alpha$:Given, $\mathrm{cos}\left(\alpha +\beta \right)=\frac{4}{5}$$⇒$ $\mathrm{sin}\left(\alpha +\beta \right)=\frac{3}{5}$ $\mathrm{sin}\left(\alpha -\beta \right)=\frac{5}{13}$$⇒$ $\mathrm{cos}\left(\alpha -\beta \right)=\frac{12}{13}$Now, we can write$2\alpha =\alpha +\beta +\alpha –\beta$Step 2. Take $"\mathrm{tan}"$ on both sides, we get$\mathrm{tan}2\alpha =\mathrm{tan}\left(\alpha +\beta +\alpha –\beta \right)$$\mathrm{tan}2\alpha =\frac{\left[\mathrm{tan}\left(\alpha +\beta \right)+\mathrm{tan}\left(\alpha –\beta \right)\right]}{\left[1–\mathrm{tan}\left(\alpha +\beta \right)\mathrm{tan}\left(\alpha –\beta \right)\right]}$ …(1) $\left[\mathbf{\because }\mathbit{t}\mathbit{a}\mathbit{n}\left(\theta +\varphi \right)\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\varphi }}{\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\theta }\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\varphi }}\right]$Also,$\begin{array}{rcl}\mathrm{tan}\left(\alpha +\beta \right)& =& \frac{\mathrm{sin}\left(\alpha +\beta \right)}{\mathrm{cos}\left(\alpha +\beta \right)}\\ & =& \frac{3/5}{4/5}\\ & =& \frac{3}{4}\end{array}$$\begin{array}{rcl}\mathrm{tan}\left(\alpha –\beta \right)& =& \frac{\mathrm{sin}\left(\alpha –\beta \right)}{\mathrm{cos}\left(\alpha –\beta \right)}\\ & =& \frac{5/13}{12/13}\\ & =& \frac{5}{12}\end{array}$Step 3. Put these values in equation (1), we get$\begin{array}{rcl}\therefore \mathrm{tan}2\alpha & =& \frac{\left(3/4\right)+\left(5/12\right)}{1–\left(3/4\right)\left(5/12\right)}\\ & =& \frac{\left(9+5\right)/12}{\left(48–15\right)/48}\\ & =& \frac{\mathbf{56}}{\mathbf{33}}\end{array}$Hence, Option ‘B’ is Correct.

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