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Popular >

f(θ)=(tan^2(θ))/(sec(θ)+1)

Solution

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

Solution

Period : 2 π

Domain : 2 πn \leq θ < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < θ < π + 2 πn or π + 2 πn < θ < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < θ < 2 π + 2 πn

Range : f (θ) < - 2 or f (θ) \geq 0

Y Intercepts : (0, 0)

Asymptotes : Vertical θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{2} + 2 πn

Extreme Points : None

+1

Interval Notation

Domain : [2 πn, \frac{π}{2} + 2 πn) \cup (\frac{π}{2} + 2 πn, π + 2 πn) \cup (π + 2 πn, \frac{3 π}{2} + 2 πn) \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn)

Range : (- \infty, - 2) \cup [0, \infty)

Solution steps

Periodicity of

\frac{tan ^{2} ( θ )}{sec ( θ ) + 1} : 2 π

Domain of

\frac{tan ^{2} ( θ )}{sec ( θ ) + 1} : 2 πn \leq θ < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < θ < π + 2 πn or π + 2 πn < θ < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < θ < 2 π + 2 πn

Range of

\frac{tan ^{2} ( θ )}{sec ( θ ) + 1} : f (θ) < - 2 or f (θ) \geq 0

Axis interception points of

\frac{tan ^{2} ( θ )}{sec ( θ ) + 1} :

Y Intercepts

: (0, 0)

Asymptotes of

\frac{tan ^{2} ( θ )}{sec ( θ ) + 1} :

Vertical

: θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{2} + 2 πn

Extreme Points of

\frac{tan ^{2} ( θ )}{sec ( θ ) + 1} :

None

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Frequently Asked Questions (FAQ)

What is the f(θ)=(tan^2(θ))/(sec(θ)+1) ?
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