What is the domain of (cos(x))/(1+tan(x)) ?

The domain of (cos(x))/(1+tan(x)) is

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domain of (cos(x))/(1+tan(x))

domain

2 πn \leq x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{3 π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < \frac{7 π}{4} + 2 πn or \frac{7 π}{4} + 2 πn < x < 2 π + 2 πn

Interval Notation

[2 πn, \frac{π}{2} + 2 πn) \cup (\frac{π}{2} + 2 πn, \frac{3 π}{4} + 2 πn) \cup (\frac{3 π}{4} + 2 πn, \frac{3 π}{2} + 2 πn) \cup (\frac{3 π}{2} + 2 πn, \frac{7 π}{4} + 2 πn) \cup (\frac{7 π}{4} + 2 πn, 2 π + 2 πn)

Solution steps

Take the denominator(s) of

\frac{cos ( x )}{1 + tan ( x )}

and compare to zero:

x = \frac{3 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Periodicity of

\frac{cos ( x )}{1 + tan ( x )} : 2 π

Combine period:

2 πn \leq x < 2 π + 2 πn

and undefined points.

The domain is

2 πn \leq x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{3 π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < \frac{7 π}{4} + 2 πn or \frac{7 π}{4} + 2 πn < x < 2 π + 2 πn

What is the domain of (cos(x))/(1+tan(x)) ?
The domain of (cos(x))/(1+tan(x)) is