How do you solve and find the value of $cot ({sin}^{- 1} (\frac{5}{6}))$ $cot(sin^-1(5/6))$ ?

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How do you solve and find the value of $cot ({sin}^{- 1} (\frac{5}{6}))$ $cot(sin^-1(5/6))$ ?

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Answer 1 · 2017-01-19T14:21:33

$cot θ = + \frac{\sqrt{11}}{5}$ $cottheta=+sqrt11/5$ .

Explanation:

Let ${sin}^{- 1} (\frac{5}{6}) = θ \Rightarrow sin θ = \frac{5}{6}, θ \in [- \frac{π}{2}, \frac{π}{2}]$ $sin^-1(5/6)=theta rArr sintheta=5/6, theta in [-pi/2,pi/2]$

But as $sin θ > 0, θ \in (0, \frac{π}{2})$ $sintheta >0, theta in (0,pi/2)$

We require the value= $cot θ$ $cottheta$

Knowing that #csc^2theta=1+cot^2theta, we have,

${cot}^{2} θ = {csc}^{2} θ - 1 = \frac{1}{{sin}^{2} θ} - 1 = \frac{36}{25} - 1 = \frac{11}{25}$ $cot^2theta=csc^2theta-1=1/sin^2theta-1=36/25-1=11/25$

$\Rightarrow cot θ = \pm \frac{\sqrt{11}}{5}$ $rArr cottheta=+-sqrt11/5$

As, $θ \in (0, \frac{π}{2}), cot θ = + \frac{\sqrt{11}}{5}$ $theta in (0,pi/2), cottheta=+sqrt11/5$ .

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How do you solve and find the value of $cot ({sin}^{- 1} (\frac{5}{6}))$ $cot(sin^-1(5/6))$ ?

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How do you solve and find the value of $cot ({sin}^{- 1} (\frac{5}{6}))$ $cot(sin^-1(5/6))$ ?