Calculate $$\frac{6 \sin(A) \sin (B) \cdot \sin (C)+\cos(A)\cdot \cos(B) \cdot \cos C)} { \cos(A+B) \cdot \cos(B+C) \cdot \cos(A+C)}$$ where $A+B+C=81$ and $\tan81=6$.
I have tried all of this question but it doesn’t work. Could you please give me some clue?
HINTS:
First suppose $A=B=0,\, C=81^\circ.$ Substituting into the given formula gives $\sec(81^\circ)$, so apparently, this is what you have to prove. Now, I did some computer experiments, and if I leave $6$ in the formula, I don't get a constant value, but if I replace $6$ by $\tan(81^\circ)$ then I do. Therefore, it appears that what we have to prove is that $$ {\tan(A+B+C)\sin{A}\sin{B}\sin{C}+\cos{A}\cos{B}\cos{C}\over\cos(A+B)\cos(B+C)\cos(C+A)}={1\over\cos(A+B+C)}$$
Cross-multiply and use the addition formulas to see if you can verify this.
EDIT
Calculation with sympy confirms that the formula is true. Good luck.
P.S.
If you know some calculus, a rather easy way to do this is to note that we need to prove $$F(A,B,C)\equiv0,$$ where $$ F(A,B,C) = \sin(A+B+C)\sin{A}\sin{B}\sin{C}+\cos(A+B+C)\cos{B}\cos{C}-\\\cos(A+B)\cos(B+C)\cos(C+A)$$
It's straightforward to compute ${\partial F\over\partial A}=0,$ which means that the value doesn't depend on $A$, so we need only verify $F(A,B,C)=)$ for one value of $A$. This is trivial when we set $A=0.$ (If you aren't acquainted with partial derivatives yet, it just means we treat $B$ and $C$ as constants for the moment, and differentiate with respect to $A$.)