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Question

Let $$\displaystyle \cos(\alpha+\beta)=\frac{4}{5}$$ and $$\displaystyle \sin(\alpha-\beta)=\frac{5}{13}$$ , where $$ 0\leq\alpha,\ \displaystyle \beta\leq\frac{\pi}{4}$$ , then $$\tan 2\alpha$$ is equal to


A
5633
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B
1912
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C
207
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D
2516
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Solution

The correct option is B $$\displaystyle \frac{56}{33}$$
$$\tan 2\alpha =\dfrac{\sin 2\alpha }{\cos 2\alpha }$$

$$\Rightarrow \tan 2\alpha =\dfrac{\sin 2\alpha }{1-2\sin ^2 2\alpha }$$

$$\sin 2\alpha =sin [(\alpha +\beta )+(\alpha -\beta )]=\sin (\alpha +\beta )\cos (\alpha -\beta )+\sin (\alpha -\beta )\cos (\alpha +\beta )$$

$$=\dfrac {3} {5} \times \dfrac{12} {13} + \dfrac{5} {13} \times \dfrac{4} {5} = \dfrac{36} {65} + \dfrac{20} {65} =\dfrac{ 56} {65 }$$

$$\cos 2\alpha =  1-2\sin ^2 2\alpha = 1-2 \left ( \dfrac{ 56} {65 } \right )^2=\dfrac{33}{65}$$

$$\tan 2\alpha =\dfrac{\frac{ 56} {65 } }{\frac{33}{65} }=\dfrac{56}{33}$$


Mathematics

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