How do you solve and find the value of $cot ({sin}^{- 1} (\frac{7}{9}))$ $cot(sin^-1(7/9))$ ?

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How do you solve and find the value of $cot ({sin}^{- 1} (\frac{7}{9}))$ $cot(sin^-1(7/9))$ ?

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Answer 1 · 2017-03-10T05:42:14

$\pm \frac{4 \sqrt{2}}{7}$ $+- (4sqrt2)/7$

Explanation:

$sin x = \frac{7}{9}$ $sin x = 7/9$
cot (arcsin x) = cot x
Use trig identity:
$1 + {cot}^{2} x = \frac{1}{{sin}^{2} x} = \frac{81}{49}$ $1 + cot^2 x = 1/(sin^2 x) = 81/49$
${cot}^{2} x = \frac{81}{49} - 1 = \frac{32}{49}$ $cot^2 x = 81/49 - 1 = 32/49$
$cot x = \pm \frac{4 \sqrt{2}}{7}$ $cot x = +- (4sqrt2)/7$ .
$sin x = \frac{7}{9}$ $sin x = 7/9$ --> x could be in Quadrant 1 or Quadrant 2. There fore,
cot x could be either negative or positive.

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How do you solve and find the value of $cot ({sin}^{- 1} (\frac{7}{9}))$ $cot(sin^-1(7/9))$ ?

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