What is the f(z)=(sec^4(z)-1)/(2+tan^2(z)) ?

The f(z)=(sec^4(z)-1)/(2+tan^2(z)) is

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f(z)=(sec^4(z)-1)/(2+tan^2(z))

f (z) = \frac{sec ^{4} ( z ) - 1}{2 + tan ^{2} ( z )}

Period : π

Domain : πn \leq z < \frac{π}{2} + πn or \frac{π}{2} + πn < z < π + πn

Range : f (z) \geq 0

X Intercepts : (πn, 0), Y Intercepts : (0, 0)

Asymptotes : Vertical z = \frac{π}{2} + πn

Extreme Points : Minimum (πn, 0)

Interval Notation

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

Range : [0, \infty)

Solution steps

Periodicity of

\frac{sec ^{4} ( z ) - 1}{2 + tan ^{2} ( z )} : π

Domain of

\frac{sec ^{4} ( z ) - 1}{2 + tan ^{2} ( z )} : πn \leq z < \frac{π}{2} + πn or \frac{π}{2} + πn < z < π + πn

Range of

\frac{sec ^{4} ( z ) - 1}{2 + tan ^{2} ( z )} : f (z) \geq 0

Axis interception points of

\frac{sec ^{4} ( z ) - 1}{2 + tan ^{2} ( z )} :

X Intercepts

: (πn, 0),

Y Intercepts

: (0, 0)

Asymptotes of

\frac{sec ^{4} ( z ) - 1}{2 + tan ^{2} ( z )} :

Vertical

: z = \frac{π}{2} + πn

Extreme Points of

\frac{sec ^{4} ( z ) - 1}{2 + tan ^{2} ( z )} :

Minimum

(πn, 0)

y = \frac{5}{2} x

y = - x^{2} + 8 x - 15

f (x) = \frac{4}{x - 3}

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

y = lo g_{7} (x)

What is the f(z)=(sec^4(z)-1)/(2+tan^2(z)) ?
The f(z)=(sec^4(z)-1)/(2+tan^2(z)) is