How do you find the domain and range of $y = ln ({tan}^{2} (x))$ $y=ln(tan^2(x))$ ?

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How do you find the domain and range of $y = ln ({tan}^{2} (x))$ $y=ln(tan^2(x))$ ?

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Answer 1 · 2018-07-29T03:34:01

Asymptotic $x ne k pi darr and ne (2k + 1)pi/2 uarr, k = 0, +-1, +-2, +-3, ...$

Explanation:

$y = ln ((tan x )^2) = 2 ln tan x$ is real when

$x ne k pi , ne (2k + 1)pi/2 and notin Q_2$ or $Q_4$ ,

to negate respectively, $sin x = 0 and cos x = 0$ .
Graph:
graph{(tan x - e^(y/2))(x)(x^2-1/4(pi)^2)(x^2-(pi)^2)(x^2-9/4(pi)^2)(x^2-4(pi)^2)(x^2-25/4(pi)^2)(x^2-9(pi)^2)=0}

Note that $y to oo, x to ( 2k + 1 )pi/2 and - oo, x to kpi$ .

It is delightful yo observe alternate asymptoticity, by sliding the graph $uarr and darr$ . As you just see, it is in hiding.

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How do you find the domain and range of $y = ln ({tan}^{2} (x))$ $y=ln(tan^2(x))$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the domain and range of y=ln(tan^2(x))y=ln(tan2(x))y=ln(tan^2(x))?

1 Answer

Explanation:

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Impact of this question

How do you find the domain and range of $y = ln ({tan}^{2} (x))$ $y=ln(tan^2(x))$ ?