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

The f(θ)=(2tan(1/2)θ)/(1+tan^2(1/2)θ) is

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f(θ)=(2tan(1/2)θ)/(1+tan^2(1/2)θ)

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

Domain : θ < - \frac{1}{tan ^{2} ( \frac{1}{2} )} or θ > - \frac{1}{tan ^{2} ( \frac{1}{2} )}

Y Intercepts : (0, 0)

Asymptotes : Vertical θ = - \frac{1}{tan ^{2} ( \frac{1}{2} )}, Horizontal y = \frac{2}{tan ( \frac{1}{2} )}

Interval Notation

Domain : (- \infty, - \frac{1}{tan ^{2} ( \frac{1}{2} )}) \cup (- \frac{1}{tan ^{2} ( \frac{1}{2} )}, \infty)

Solution steps

Domain of

\frac{2 tan ( \frac{1}{2} ) θ}{1 + tan ^{2} ( \frac{1}{2} ) θ} : θ < - \frac{1}{tan ^{2} ( \frac{1}{2} )} or θ > - \frac{1}{tan ^{2} ( \frac{1}{2} )}

Axis interception points of

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

Y Intercepts

: (0, 0)

Asymptotes of

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

Vertical

: θ = - \frac{1}{tan ^{2} ( \frac{1}{2} )},

Horizontal

: y = \frac{2}{tan ( \frac{1}{2} )}

y = \frac{4}{5} x + 6

g (x) = \frac{x}{x - 1}

f (x) = x + 1, 0 \leq x \leq 3

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

f (x) = 1 + e^{2 x}

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