How do you evaluate $tan ({sin}^{- 1} (- \frac{1}{6}))$ $tan(sin^-1(-1/6))$ ?

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How do you evaluate $tan ({sin}^{- 1} (- \frac{1}{6}))$ $tan(sin^-1(-1/6))$ ?

1 Answer

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Answer 1 · 2016-05-30T16:21:10

$- \frac{1}{\sqrt{35}}$ $-1/sqrt 35$ .

Explanation:

Let $a = {sin}^{- 1} (- \frac{1}{6})$ $a = sin^(-1) (-1/6)$ . Then, $sin a = - \frac{1}{6} < 0$ $sin a = -1/6 < 0$ . $a$ $a$ is in the 3rd quadrant or in the 4th. On the other hand, he "principal branch" of the inverse sine corresponds to an angle in the first or fourth quadrant, not the third. So we pick the fourth quadrant angle, and $cos a = + \frac{\sqrt{35}}{6}$ $cos a = +sqrt 35/6$ .

The given expression $= tan a = \frac{sin a}{cos a} = - \frac{1}{\sqrt{35}}$ $= tan a = sin a/cos a=-1/sqrt 35$ .

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How do you evaluate $tan ({sin}^{- 1} (- \frac{1}{6}))$ $tan(sin^-1(-1/6))$ ?

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