Is $\sqrt{1-\tan^2(x)}$ continuous? I think how $\sqrt{x}$ is continuous in $[0,\infty) $ then  $\sqrt{1-\tan^2(x)}$ is continuous when $1-\tan^2(x)\geq0$ but wolfram says that is not continuous so I am confused
 A: This is a common misconception about continuity and the statement needs to be clearly defined. On its domain, the function $f(x) = \sqrt{1-\tan^2(x)}$ is continuous. No more than that.
By definition of continuity, it does not make sense to ask whether a function is continuous in points where it is not defined.
A: tanx is not defined at odd multiples of π/2 so the function would also be not defined at those points. Hence the function would not be continuous at odd multiples of π/2.
A: Let $f(x)=\sqrt{1-\tan^2(x)}.$

*

*$f$ is a continuous function. This is because formally, a
continuous function is one that is continuous on its domain.
$f$  is undefined on $\left(\frac{4k+1}4,\frac{4k+3}4\right)$ for
each $n\in\mathbb Z;$ on these intervals, the continuity of $f$ is
not a meaningful concept.


*However, the claim “$f$ is not continuous” can be charitably
interpreted to implicitly mean $$“f \text{ is not continuous on
}\mathbb R”,$$ which, although unmeaningful, can in turn be
charitably interpreted either as being vacuously true, or as $$“\text{ it is not that }f \text{ is continuous on }  \mathbb R”,$$
which is technically true since $f$ is actually neither continuous or
discontinuous on $\mathbb R.$
I'm just bending over backwards to bridge the pre-calculus and formal notions of continuity.
