Solve $\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}=4x^2.$ I just need help to finetune this solution. Only having a correct answer is not sufficient to comb home 5/5 points on a problem like this. Thoughts on improvements on stringency? Is there any logical fallacy or ambiguity? Any input is very welcome!

Solution: The domain for $\arctan{x}$ is $\mathbb{R},$ thus the function $\sin{(\arctan{x})}$ is defined over the entire $\mathbb{R}$. However, the restricting factor comes from the fact that the domain for $\arcsin{x}$ is $[-1,1]$ which implies that the domain for $\tan{(\arcsin{x})}$ is $[-\arctan{1},\arctan{1}]=[-\frac{\pi}{4},\frac{\pi}{4}].$ We can thus conclude that if there exists a solution $x$ to the above equation, then $x\in[-\frac{\pi}{4},\frac{\pi}{4}]$.
First I note that $$\sin{b}=\pm\frac{\tan{b}}{\sqrt{\tan^2{b}+1}}\Rightarrow\sin^2{(\arctan{x})}=\left(\pm\frac{\tan{(\arctan{x})}}{\sqrt{\tan^2{(\arctan{x})}+1}}\right)^2=\frac{x^2}{x^2+1}.$$
Secondly; $$\tan{c}=\pm\frac{\sin{c}}{\sqrt{1-\sin^2{c}}}\Rightarrow \tan^2{\arcsin{x}}=\left(\pm\frac{\sin{(\arcsin{x})}}{\sqrt{1-\sin^2{(\arcsin{x})}}}\right)^2=\frac{x^2}{1-x^2}.$$
So the equation becomes $$\frac{x^2+1-1+x^2}{x^2}=\frac{2x^2}{x^2} =4x^2\Longleftrightarrow x=\pm\frac{1}{\sqrt{2}}.$$
What's left to do now is to show that these $x:$s both lie in the desired interval. So it boils down to showing that $\frac{1}{\sqrt{2}}\leq\frac{\pi}{4}$. This can be done by multiplying the inequality by $\sqrt{2}/\sqrt{2}$: $$\frac{\sqrt{2}}{4}\leq \frac{\pi}{4}\Longleftrightarrow \sqrt{2}\leq \pi,$$
which clearly is true.
 A: The allowed values for $x$ are in $[-1,1]$, because of $\arcsin x$, but $0$ should also be excluded. Also $-1$ and $1$ must be excluded because of $\tan\arcsin x$.
(Note: your $x\in[-\pi/4,\pi/4]$ is wrong and the probable cause for the low grade.)
The equation remains the same if we change $x$ into $-x$, so we can limit ourselves to $x\in(0,1)$. For every root we find, also its negative will be a root.
If $y=\sin\arctan x$, then $\arctan x=\arcsin y$ and so $x=\tan\arcsin y$ so
$$
x^2=\frac{\sin^2\arcsin y}{1-\sin^2\arcsin y}=\frac{y^2}{1-y^2}
$$
whence
$$
y^2=\frac{x^2}{1+x^2}
$$
If $z=\tan\arcsin x$, then
$$
z^2=\frac{\sin^2\arcsin x}{1-\sin^2\arcsin x}=\frac{x^2}{1-x^2}
$$
So your equation $\frac{1}{y^2}-\frac{1}{z^2}=4x^2$ becomes
$$
\frac{1+x^2}{x^2}-\frac{1-x^2}{x^2}=4x^2
$$
that is,
$$
2x^2=4x^4
$$
and, since $x\ne0$, $x^2=\frac{1}{2}$. The only positive root is
$$
x=\frac{1}{\sqrt{2}}
$$
which indeed belongs to $(0,1)$. Thus the solutions are
$$
\frac{1}{\sqrt{2}}\qquad\text{and}\qquad{-}\frac{1}{\sqrt{2}}
$$
A: Taking derivative of LHS we have $$\frac{d}{dx}\left(\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}\right)=0$$
Therefore $$\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}=\text{ constant }$$
Calculation for $x=\dfrac{1}{\sqrt3}$ gives
$$\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}=2$$
Now the points of the parabola $y=4x^2$ with $2=4x^2$ are $\color{red}{x=\pm\dfrac{1}{\sqrt2}}$
