What is the domain of f(x)= 1/(1-tan(x)) ?

The domain of f(x)= 1/(1-tan(x)) is pin<= x< pi/4+pin\lor pi/4+pin<x< pi/2+pin\lor pi/2+pin<x<pi+pin

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domain of f(x)= 1/(1-tan(x))

domain

f (x) = \frac{1}{1 - tan ( x )}

πn \leq x < \frac{π}{4} + πn or \frac{π}{4} + πn < x < \frac{π}{2} + πn or \frac{π}{2} + πn < x < π + πn

Interval Notation

[πn, \frac{π}{4} + πn) \cup (\frac{π}{4} + πn, \frac{π}{2} + πn) \cup (\frac{π}{2} + πn, π + πn)

Solution steps

Take the denominator(s) of

\frac{1}{1 - tan ( x )}

and compare to zero:

x = \frac{π}{4} + πn, x = \frac{π}{2} + πn

Periodicity of

\frac{1}{1 - tan ( x )} : π

Combine period:

πn \leq x < π + πn

and undefined points.

The domain is

πn \leq x < \frac{π}{4} + πn or \frac{π}{4} + πn < x < \frac{π}{2} + πn or \frac{π}{2} + πn < x < π + πn

lo g_{10} (x + 2) - 3

f (x) = \frac{1}{2} x

f (x) = x^{2} - 2 x + 3

(1, 3), (- 2, 9)

2 x^{2} + 16 x + 12

What is the domain of f(x)= 1/(1-tan(x)) ?
The domain of f(x)= 1/(1-tan(x)) is pin<= x< pi/4+pin\lor pi/4+pin<x< pi/2+pin\lor pi/2+pin<x<pi+pin