1. Use the given angle ranges

Since $0 \le \alpha,\beta \le \dfrac{\pi}{4}$, we have:

• $0 \le \alpha+\beta \le \dfrac{\pi}{2}$
• $-\dfrac{\pi}{4} \le \alpha-\beta \le \dfrac{\pi}{4}$

So the relevant sine and cosine values will be taken with positive cosine, and for $\alpha-\beta$ the cosine is also positive.

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2. Find $\sin(\alpha+\beta)$

Given

$$\cos(\alpha+\beta)=\frac45$$

and $\alpha+\beta \in \left[0,\frac\pi2\right]$, so

$$\sin(\alpha+\beta)=\sqrt{1-\left(\frac45\right)^2} =\sqrt{1-\frac{16}{25}} =\frac35.$$

Thus,

$$\sin(\alpha+\beta)=\frac35, \qquad \cos(\alpha+\beta)=\frac45.$$

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3. Find $\cos(\alpha-\beta)$

Given

$$\sin(\alpha-\beta)=\frac5{13}.$$

Since $\alpha-\beta \in \left[-\frac\pi4,\frac\pi4\right]$, we have $\cos(\alpha-\beta)>0$. Therefore,

$$\cos(\alpha-\beta)=\sqrt{1-\left(\frac5{13}\right)^2} =\sqrt{1-\frac{25}{169}} =\frac{12}{13}.$$

So,

$$\sin(\alpha-\beta)=\frac5{13}, \qquad \cos(\alpha-\beta)=\frac{12}{13}.$$

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4. Express $2\alpha$ in terms of these angles

Notice that

$$2\alpha=(\alpha+\beta)+(\alpha-\beta).$$

Hence,

$$\tan 2\alpha=\tan\big((\alpha+\beta)+(\alpha-\beta)\big).$$

Let

$$X=\alpha+\beta, \qquad Y=\alpha-\beta.$$

Then

$$\tan 2\alpha=\tan(X+Y)=\frac{\tan X+\tan Y}{1-\tan X\tan Y}.$$

Now,

$$\tan X=\frac{\sin X}{\cos X}=\frac{3/5}{4/5}=\frac34,$$ $$\tan Y=\frac{\sin Y}{\cos Y}=\frac{5/13}{12/13}=\frac5{12}.$$

Therefore,

$$\tan 2\alpha=\frac{\frac34+\frac5{12}}{1-\frac34\cdot\frac5{12}}.$$

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5. Simplify

First, the numerator:

$$\frac34+\frac5{12}=\frac9{12}+\frac5{12}=\frac{14}{12}=\frac76.$$

Now, the denominator:

$$1-\frac34\cdot\frac5{12}=1-\frac{15}{48}=1-\frac5{16}=\frac{11}{16}.$$

Thus,

$$\tan 2\alpha=\frac{7/6}{11/16}=\frac76\cdot\frac{16}{11}=\frac{56}{33}.$$

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6. Match with the options

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

which corresponds to Option A.

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7. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.