What Is the Value of $\tan^2 a + 2\tan^2 b$ Given the Trigonometric Condition?

[anemone]

Evaluate $\tan^2 a+2\tan^2 b$ if $a$ and $b$ satisfy the trigonometric equality $2\sin a \sin b+3\cos b+6\cos a \sin b=7$.
This is an unsolved problem I found @ AOPS.

 

[Opalg]

Evaluate $\tan^2 a+2\tan^2 b$ if $a$ and $b$ satisfy the trigonometric equality $2\sin a \sin b+3\cos b+6\cos a \sin b=7$.

[sp]Outline solution: Write the equation as $2\sin a + 6\cos a = \dfrac{7 - 3\cos b}{\sin b}$. The maximum of the absolute value of the left side is $\sqrt{40}$, occurring when $\cos a = 3\sqrt{40}/20$ (and therefore $\tan a = \pm1/3$). The minimum of the absolute value of the right side is also $\sqrt{40}$, occurring when $\cos b = 1/3$ (and therefore $\tan b = \pm\sqrt{40}/3$). So the equation only holds when both those conditions are satisfied, in other words $\tan^2 a+2\tan^2 b = \dfrac19 + \dfrac{80}9 = 9.$

[DESMOS=-10,10,-10,10]2\sin x\ +\ 6\cos x;\frac{\left(7\ -\ 3\cos x\right)}{\sin x}[/DESMOS]
[/sp]

 

[anemone]

Thanks a ton for your constant participation to my "Unsolved Challenge" and of course, for your insightful and beautiful solutions, Opalg!

 

[topsquark]

Spoiler

If I understand what you are doing then that would mean the max and mins of the relevant graphs are the same, thus the LHS and RHS overlap at a point. But the graph shows that that doesn't happen??

-Dan

 

[MarkFL]

Spoiler

If I understand what you are doing then that would mean the max and mins of the relevant graphs are the same, thus the LHS and RHS overlap at a point. But the graph shows that that doesn't happen??

-Dan

Spoiler

The graph, and supporting algebra, suggests that there is at least one point $(a,b)$ where $a\ne b$ such that:

\(\displaystyle 2\sin(a)+6\cos(a)=7\csc(b)-3\cot(b)\)

 

[Opalg]

Spoiler

If I understand what you are doing then that would mean the max and mins of the relevant graphs are the same, thus the LHS and RHS overlap at a point. But the graph shows that that doesn't happen??

-Dan

[sp]That had me fooled too, for a long time. I assumed it meant that there were no solutions to the problem. The explanation (as Mark points out) is that the graph shows two functions of a single variable $x$, but the problem refers to two separate variables $a$ and $b$. So it does not matter that the functions attain their extreme values at different points.[/sp]