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

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

Working : 
$$B = \cot 66^{\circ} \cot78^{\circ} = 1- \frac{\tan24^{\circ}+\tan18^{\circ}}{\tan42^{\circ}}$$
$$A= \tan6^{\circ} \tan42^{\circ} = 1- \frac{\tan6^{\circ} +\tan42^{\circ}}{\tan48^{\circ}}$$
but it seems  this is the wrong way of doing this...please suggest. Thanks!
 A: 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
A: 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.
A: $$A=Tan(6^\circ)\,Tan(42^\circ)=Tan(2.3^\circ)Tan(45^\circ-3^\circ)=\frac{2Tan(3^\circ)}{1-Tan^2(3^\circ)}\frac{1-Tan(3^\circ)}{1+Tan(3^\circ)}=\frac{2Tan(3^\circ)}{\left(1+Tan(3^\circ)\right)^2}$$
$\:$
$$ B=Tan(24^\circ)\,Tan(12^\circ)=Tan(2.12^\circ)Tan(12^\circ)=\frac{2Tan^2(12^\circ)}{1-Tan^2(12^\circ)} \implies$$
$\:$
$$Tan(12^\circ)=\sqrt{\frac{B}{B+2}}$$ But $$Tan(12^\circ)=Tan(15^\circ-3^\circ)=\frac{(2-\sqrt{3})-Tan(3^\circ)}{(2-\sqrt{3})+Tan(3^\circ)}\implies$$
$$Tan(3^\circ)=\left(\frac{\sqrt{B+2}-\sqrt{B}}{\sqrt{B+2}+\sqrt{B}}\right)(2-\sqrt{3})$$  Substitute $\,$$Tan(3^\circ)$ above in the Expression of $A$ to get the Required Relation 
