You can use FunctionDomain and FunctionRange:

FunctionDomain[Tan[x], x]

1/2 + x/π ∉ Integers

FunctionRange[Tan[x],x,y]

True

FunctionRange[Sin[x], x, y]

-1 <= y <= 1

Update on questions in comments:

• how the answer $1/2 + x/π ∉ \mathbb{Z}$ is related to the right result $ x ≠ π / 2 + k π $?

The two expressions are equivalent: Move $\Pi/2$ to the lhs and divide both sides of the second expression by $\Pi$ to get $ x/π - 1/2 ≠ k $ ($k$ integer).

• Why does Mathematica return the first expression (not the second) as the answer?

The first one is simpler by LeafCount:

1/2 + x/π ∉ Integers // LeafCount

11

ForAll[k, Element[k, Integers], x != k + π/2] // LeafCount

14

You can use FunctionDomain and FunctionRange:

FunctionDomain[Tan[x], x]

1/2 + x/π ∉ Integers

FunctionRange[Tan[x],x,y]

True

FunctionRange[Sin[x], x, y]

-1 <= y <= 1

Update on questions in comments:

• how the answer $1/2 + x/π ∉ \mathbb{Z}$ is related to the right result $ x ≠ π / 2 + k π $?

The two expressions are equivalent: Move $\pi/2$ to the lhs and divide both sides of the second expression by $\pi$ to get $ x/π - 1/2 ≠ k $ ($k$ integer).

• Why does Mathematica return the first expression (not the second) as the answer?

The first one is simpler by LeafCount:

1/2 + x/π ∉ Integers // LeafCount

11

ForAll[k, Element[k, Integers], x != k + π/2] // LeafCount

14

You can use FunctionDomain and FunctionRange:

FunctionDomain[Tan[x], x]

1/2 + x/π ∉ Integers

FunctionRange[Tan[x],x,y]

True

FunctionRange[Sin[x], x, y]

-1 <= y <= 1

You can use FunctionDomain and FunctionRange:

FunctionDomain[Tan[x], x]

1/2 + x/π ∉ Integers

FunctionRange[Tan[x],x,y]

True

FunctionRange[Sin[x], x, y]

-1 <= y <= 1

Update on questions in comments:

• how the answer $1/2 + x/π ∉ \mathbb{Z}$ is related to the right result $ x ≠ π / 2 + k π $?

The two expressions are equivalent: Move $\Pi/2$ to the lhs and divide both sides of the second expression by $\Pi$ to get $ x/π - 1/2 ≠ k $ ($k$ integer).

• Why does Mathematica return the first expression (not the second) as the answer?

The first one is simpler by LeafCount:

1/2 + x/π ∉ Integers // LeafCount

11

ForAll[k, Element[k, Integers], x != k + π/2] // LeafCount

14