What is the domain of 1/(sec^2(x)) ?

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

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domain of 1/(sec^2(x))

domain

\frac{1}{sec ^{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}{sec ^{2} ( x )}

and compare to zero:

x = \frac{π}{2} + πn

Periodicity of

\frac{1}{sec ^{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{3 x}{4 - x}

f (x) = - 0.05 x^{2} + 10

f (x) = \frac{x ^{2} - 64}{x ( x - 8 )}

f (x) = 3 + x - 2

3 x - 2 y = 16

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