Ortega Adkins
05/01/2024 · Senior High School
Find a simplified expression for \( \tan \left(\sin ^{-1}\left(\frac{b}{10}\right)\right) \) \( \square \)
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Tutor-Verified Answer
Step-by-step Solution
To find a simplified expression for \( \tan \left(\sin^{-1}\left(\frac{b}{10}\right)\right) \), we start by letting
\[
\theta = \sin^{-1}\left(\frac{b}{10}\right).
\]
This means that
\[
\sin \theta = \frac{b}{10}.
\]
We want to find \( \tan \theta \). Recall that
\[
\tan \theta = \frac{\sin \theta}{\cos \theta}.
\]
We already have \( \sin \theta \), so we need to find \( \cos \theta \). We can use the Pythagorean identity:
\[
\sin^2 \theta + \cos^2 \theta = 1.
\]
Substituting \( \sin \theta \):
\[
\left(\frac{b}{10}\right)^2 + \cos^2 \theta = 1.
\]
This simplifies to:
\[
\frac{b^2}{100} + \cos^2 \theta = 1.
\]
Rearranging gives:
\[
\cos^2 \theta = 1 - \frac{b^2}{100} = \frac{100 - b^2}{100}.
\]
Taking the square root, we find:
\[
\cos \theta = \sqrt{\frac{100 - b^2}{100}} = \frac{\sqrt{100 - b^2}}{10}.
\]
Now we can substitute \( \sin \theta \) and \( \cos \theta \) into the expression for \( \tan \theta \):
\[
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{b}{10}}{\frac{\sqrt{100 - b^2}}{10}}.
\]
The \( \frac{1}{10} \) cancels out:
\[
\tan \theta = \frac{b}{\sqrt{100 - b^2}}.
\]
Thus, the simplified expression for \( \tan \left(\sin^{-1}\left(\frac{b}{10}\right)\right) \) is
\[
\boxed{\frac{b}{\sqrt{100 - b^2}}}.
\]
Quick Answer
The simplified expression is \( \frac{b}{\sqrt{100 - b^2}} \).
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