Question: If $\cos (A+B)=\dfrac {3}{5}$ and $\cos (A-B)=\dfrac {4}{5}$, prove that: $A=45°$.

My Attempt:
$$\cos (A+B)-\cos (A-B)=\dfrac {3}{5} - \dfrac {4}{5}$$
$$2\sin \dfrac {A+B+A-B}{2}\cdot \sin \dfrac {A-B-B-A}{2}=\dfrac {-1}{5}$$
$$\sin A\cdot \sin B=\dfrac {1}{5}$$

How do I continue? ?

Answer: 
As written, the problem does not have a uniquely determined solution. For example, if $A,B$ is a solution then $-A,-B$ is also a solution.

Hint: $\;\sin(A+B)=\pm\sqrt{1 - \cos^2(A+B)}=\pm \cfrac{4}{5}\,$, and similarly $\sin(A-B)=\pm \cfrac{3}{5}\,$. Then:

$$
\begin{align}
\sin(2A) &= \sin\big((A+B)+(A-B)\big) \\
 &= \sin(A+B)\cos(A-B) + \sin(A-B)\cos(A+B) \\
 &= \pm \cfrac {16}{25} \pm \cfrac{9}{25} \;\;\in\;\; \{ -1, \cfrac{-7}{25}, \cfrac{7}{25}, 1 \}
\end{align}
$$

If $\sin(2A) = 1$ then $A=\cfrac{\pi}{4}=45^\circ\,$. To exclude the other cases, additional information would be needed about the angles $A,B$.
Are there any additional restrictions on the angles $A$ and $B$? Because as currently stated, this claim isn't even true. Even if $A$ and $B$ are required to be in the first quadrant, then this is still not true. One of them does have to be $45^{\circ}$, but it can be either $A$ or $B$. That is due to the complete symmetry of the question: since $\cos(A+B)=\cos(B+A)$ and $\cos(A-B)=\cos(B-A)$, whenever $(A,B)$ is a solution, so is $(B,A)$.

The only solution within the first quadrant is the following pair of angles:

$$\varphi=\frac{\pi}{4}=45^{\circ} \quad \text{and} \quad \psi=\arcsin\frac{1}{5\sqrt{2}},$$

but they can be assigned to $A$ and $B$ in either order.

---

Here's an outline of a possible solution, skipping some details. First, notice that with the given values

$$\cos^2(A+B)+\cos^2(A-B)=1.$$

Comparing that with the fundamental trigonometric identity

$$\cos^2(A+B)+\sin^2(A+B)=1,$$

we get that

$$\cos^2(A-B)=\sin^2(A+B) \quad \Longrightarrow \quad \cos(A-B)=\pm\sin(A+B).$$

So we have two cases:

1. Case one: $\cos(A-B)=\sin(A+B)$. Since
$$\sin(A+B)=\cos\left(\frac{\pi}{2}-(A+B)\right)=\cos\left(\frac{\pi}{2}-A-B\right),$$
we have
$$\cos(A-B)=\cos\left(\frac{\pi}{2}-A-B\right),$$
which implies
$$\color{red}{A-B=\pm\left(\frac{\pi}{2}-A-B\right)+2\pi n}.$$
2. Case two: $\cos(A-B)=-\sin(A+B)$. Since
$$-\sin(A+B)=\cos\left(\frac{\pi}{2}+(A+B)\right)=\cos\left(\frac{\pi}{2}+A+B\right),$$
we have
$$\cos(A-B)=\cos\left(\frac{\pi}{2}+A+B\right),$$
which implies
$$\color{red}{A-B=\pm\left(\frac{\pi}{2}+A+B\right)+2\pi n}.$$

Solving the four red equations will give us the complete answer to the problem. One of the angles will have to be $\displaystyle \pm\frac{\pi}{4}=\pm45^{\circ}$, but it can be either $A$ or $B$.