If $\arcsin(x^2+y^2)+\arctan(4y^2-1)+\operatorname{arcsec}(x)=a$ then $a$ belongs to
p) $\{0\}$
q) $(\frac\pi2,\frac{3\pi}2)$
r) $[0,\frac\pi2)$
s) $\mathrm R$
t) $\mathrm R-\{0\}$
My Attempt:
$\operatorname{arcsec}(x)$ is defined when $x^2\ge1$, therefore for $\arcsin(x^2+y^2)$ to define, $y^2=0, x^2=1$, so,
$$a=\frac\pi2-\frac\pi4+0=\frac\pi4;\;x=1$$
Or, $$a=\frac\pi2-\frac\pi4+\pi=\frac{5\pi}4;\;x=-1$$
Therefore, my answer is $a\in\{\frac\pi4,\frac{5\pi}4\}$
But the answer given is p), q), r), s).
Can you share your thoughts? Thanks.