Range of \(\displaystyle{y}={{\cot}^{{-{1}}}{x}}{{\cot}^{{-{1}}}{\left(-{x}\right)}}\) Attempt: Let \(\displaystyle{{\cot}^{{-{1}}}{x}}=\theta\) .

Dexter Odom

Dexter Odom

Answered question

2022-04-01

Range of y=cot1xcot1(x)
Attempt: Let cot1x=θ . Then y=θ(πθ) ;since θ(0,π).Then θ2πθ+y=0.For θ to be real, yπ24
Thus 0<yπ24

Answer & Explanation

Denise Daniel

Denise Daniel

Beginner2022-04-02Added 8 answers

The step you have inquired about is correct; we may find the values of y for which θ2πθ+y=0 has (at least one) real solution through its discriminant π241y , which has to be non-negative. This gives your inequality yπ24.
Another way to solve the problem is to use the identity cot1x=π2tan1x to rewrite the given function as
y=(π2tan1x)(π2tan1(x))
Since tan1x is an odd function:
y=(π2tan1x)(π2+tan1x)=π24(tan1x)2
Since tan1x has range (π2,π2), (tan1x)2 has range [0,π24), giving the range of y as (0,π24]

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