If $A = \tan6^{\circ} \tan42^{\circ},~~B = \cot 66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$

Using

$2\cos A\cos B=\cos(A-B)+\cos(A+B)$ and $2\sin A\sin B=\cos(A-B)-\cos(A+B),$

$$A=\frac{\sin 6^\circ\cdot \sin 42^\circ}{\cos 6^\circ\cdot \cos 42^\circ}=\frac{\cos36^\circ-\cos48^\circ}{\cos36^\circ+\cos48^\circ}$$

Applying Componendo and dividendo, $$\frac{1+A}{1-A}=\frac{\cos36^\circ}{\cos48^\circ}$$

Similarly, Using

$2\sin A\cos B=\sin(A+B)+\sin(A-B)$ and $2\cos A\sin B=\sin(A+B)-\sin(A-B),$

$$B=\frac{\cos66^\circ \cos72^\circ}{\sin66^\circ \sin72^\circ}=\frac{\cos66^\circ \sin18^\circ}{\sin66^\circ \cos18^\circ}=\frac{\sin84^\circ-\sin48^\circ}{\sin84^\circ+\sin48^\circ}$$

Applying Componendo and dividendo, $$\frac{1+B}{1-B}=\frac{\sin84^\circ}{\sin48^\circ}$$

$$\implies \frac{1+A}{1-A}\cdot\frac{1+B}{1-B}=\frac{\sin84^\circ\cdot \cos36^\circ}{\sin48^\circ\cdot \cos48^\circ}=\frac{2\sin84^\circ\cdot \cos36^\circ}{\sin(2\cdot48)^\circ}=2\cos36^\circ$$ as $\sin96^\circ=\sin(180-96)^\circ=\sin84^\circ$

Now $\cos36^\circ$ can be found here

First, note that $A \approx 0.0946362785$, $\ \ B \approx 0.0946362785$.

Now, we will prove that $\ \ \ \Large{A=B.}$

a).

$$ \dfrac{A}{B} = \dfrac {\sin 6^\circ \sin 42^\circ} {\cos 6^\circ \cos 42^\circ} \cdot \dfrac {\sin 66^\circ \sin 78^\circ} {\cos 66^\circ \cos 78^\circ} = \dfrac {\bigl( 2 \sin 6^\circ \sin 66^\circ \bigr) \cdot \bigl( 2 \sin 42^\circ \sin 78^\circ \bigr)} {\bigl( 2 \cos 6^\circ \cos 42^\circ \bigr) \cdot \bigl( 2 \cos 66^\circ \cos 78^\circ \bigr)}. \tag{1} $$

b). Applying formulas

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

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

$ \ \ \ \ \ (1) \implies$ $$ \dfrac{A}{B} = \dfrac {\bigl( \cos 60^\circ - \cos72^\circ \bigr) \cdot \bigl( \cos 36^\circ - \cos 120^\circ \bigr)} {\bigl( \cos 60^\circ + \cos72^\circ \bigr) \cdot \bigl( \cos 36^\circ + \cos 120^\circ \bigr)} = \dfrac {\bigl( \frac{1}{2} - \cos72^\circ \bigr) \cdot \bigl( \cos 36^\circ + \frac{1}{2} \bigr)} {\bigl( \frac{1}{2} + \cos72^\circ \bigr) \cdot \bigl( \cos 36^\circ - \frac{1}{2} \bigr)}. \tag{2} $$

c). It is known, that $\ \ \ \ \cos72^\circ = \sin 18^\circ = \frac{1}{4}(\sqrt{5}-1)$,

so $\ \ \ \ \ \ \cos 36^\circ = 1 - 2(\sin 18^\circ)^2 = \frac{8}{8} - \frac{1}{8}(5-2\sqrt{5}+1) = \frac{1}{4}(\sqrt{5}+1)$,

and $\ \ \ \ \cos 72^\circ \cos 36^\circ = \frac{1}{16}(\sqrt{5}-1)(\sqrt{5}+1) = \frac{4}{16}=\frac{1}{4}$.

Hence (2) $\implies$

$$ \dfrac{A}{B} = \dfrac {\frac{1}{2}\cos 36^\circ - \cos 72^\circ \cos 36^\circ +\frac{1}{4}-\frac{1}{2}\cos 72^\circ} {\frac{1}{2}\cos 36^\circ +\cos 72^\circ \cos 36^\circ -\frac{1}{4}-\frac{1}{2}\cos 72^\circ } = \dfrac {\frac{1}{2}(\cos 36^\circ - \cos 72^\circ)} {\frac{1}{2}(\cos 36^\circ - \cos 72^\circ) } =\Large{1}. \tag{3} $$

Proved.