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I cannot figure out how do I get the domain for function $\tan(x^2).$ There is a square function and a tangent function. It should be all real numbers except $(\pi+k\pi)/2$ but I think the exception must be different because of square function, I just don't know what it does.

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  • $\begingroup$ So, you don't want $x^2$ to be any of those values... (You're off a bit with those, you want odd multiples of $\pi/2$.) $\endgroup$ – David Mitra Oct 25 '19 at 17:31
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Since the domain for $\tan (x)$ is $x\neq \frac{\pi}2+k\pi$ therefore for $\tan (x^2)$ we need

$$x^2\neq \frac{\pi}2+k\pi \implies x \neq \pm \sqrt{\frac{\pi}2+|k|\pi}$$

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