Solving trigonometrics functions/equations My niece ask me to help her with a school assignment, but I can't identify what type of equation are we solving.
For example:
$\DeclareMathOperator{\tg}{tg}$
$$\sin\alpha=\frac{8\sqrt{11}}{9}$$
or:
$$\tan\alpha=\frac{2\sqrt{5}}{12}$$
The notes she have are these, for example:
$$\sin\alpha=\frac{1}{\csc\alpha}$$
and:
$$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$$
I'm trying to browse for help, but I don't know wath to search for. Where to begin.

I was able to talk with a classmate of hers, she explain me this:
In each problem, we need to use all the needed functions (Right angled triangle definitions).
In the first problem:
$$\sin\alpha=\frac{8\sqrt{11}}{9}$$
We need to solve the $\cos$ then $\sec$ then $\csc$ then $\tan$ then $\cot$. Does this makes any sense?
 A: The first two are a slightly different kind of question than the bottom two. The first are basically asking "is (value on the right) anywhere in the range of (trig function on the left)?"
For $$\sin{\alpha} = \frac{8\sqrt{11}}{9}$$ there is no $\alpha$ that satisfies the equation since the range of $\sin$ is $[-1, 1]$, and the right-hand value is bigger than that.
For $$\tan{\alpha} = \frac{2\sqrt{5}}{12}$$ there is an answer, since $\tan$'s range is (almost) everything. The answer will equal $\tan^{-1}(\frac{2\sqrt{5}}{12})$.
The latter two you can view as being about whether the graphs of certain trig functions cross each other or not. Actually, though $$\sin{\alpha} = \frac{1}{\cos{\alpha}}$$ is amenable to a range argument as well. The range of $\cos{\alpha}$ is $[-1, 1]$; what range do we get if we take the reciprocal of every number in there? How does it compare to the range of $\sin{\alpha}$? (They do both attain $1$ and $-1$ -- you have to show that they don't intersect there, which shouldn't be too hard).
As for the last problem, remember that $\frac{\sin{\alpha}}{\cos{\alpha}} = \tan{\alpha}$. So
$$
\begin{aligned}
\tan{\alpha} &= \sqrt{\frac{(\sin{\alpha})^2}{(\cos{\alpha})^2}}\\
&= \sqrt{\left(\frac{\sin{\alpha}}{\cos{\alpha}}\right)^2}\\
&= \sqrt{(\tan{\alpha})^2}\\
&= |\tan{\alpha}|\\
\end{aligned}
$$
(remember the absolute value). So the problem is just asking for what $\alpha$ it's true that $\tan{\alpha} = |\tan{\alpha}|$.
