Use a right triangle to write the following expression as an algebraic expression. Assume that x is positive and in the domain of the given inverse trigonometric function. Given: displaystyle tan{{left({{cos}^{ -{{1}}}{5}}{x}right)}}=?

Question
Use a right triangle to write the following expression as an algebraic expression. Assume that x is positive and in the domain of the given inverse trigonometric function.
Given:
\(\displaystyle \tan{{\left({{\cos}^{ -{{1}}}{5}}{x}\right)}}=?\)

Answers (1)

2021-02-26
Step 1
Let,
\(\displaystyle{a}={{\cos}^{ -{{1}}}{5}}{x}\)
Then, \(\displaystyle \cos{{a}}={5}{x}\)
Therefore,
\(\displaystyle{{\sin}^{2}{a}}+{{\cos}^{2}{a}}={1}\)
\(\displaystyle{{\sin}^{2}{a}}={1}-{{\cos}^{2}{a}}\)
\(\displaystyle \sin{{a}}=\pm\sqrt{{{1}-{{\cos}^{2}{a}}}}\)
\(\displaystyle \sin{{a}}=\pm\sqrt{{{1}-{\left({5}{x}\right)}^{2}}}\)
\(\displaystyle \sin{{a}}=\pm\sqrt{{{1}-{25}{x}^{2}}}\)
Step 2
Hence,
\(\displaystyle \tan{{\left({{\cos}^{ -{{1}}}{5}}{x}\right)}}= \tan{{a}}\)
\(\displaystyle=\frac{{ \sin{{a}}}}{{ \cos{{a}}}}\)
\(\displaystyle \tan{{\left({{\cos}^{ -{{1}}}{5}}{x}\right)}}=\pm\frac{{\sqrt{{{1}-{25}{x}^{2}}}}}{{{5}{x}}}\)
0

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