What is the domain of f(x)= 1/(1+tan^2(x)) ?

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

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domain of f(x)= 1/(1+tan^2(x))

domain

f (x) = \frac{1}{1 + tan ^{2} ( x )}

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

Interval Notation

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

Solution steps

Take the denominator(s) of

\frac{1}{1 + tan ^{2} ( x )}

and compare to zero:

x = \frac{π}{2} + πn

Periodicity of

\frac{1}{1 + tan ^{2} ( x )} : π

Combine period:

πn \leq x < π + πn

and undefined points.

The domain is

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

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

y = \frac{1}{5} x + 5, (8, - 3)

f (x) = 2 x + 12

f (x) = - \frac{4}{5} x + 2

f (x) = 9 - 3 x^{2}

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