How do you find the exact value for $tan ({sin}^{- 1} (- \frac{4}{5}))$ $tan (sin^-1(-4/5))$ ?

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Answer 1 · 2016-03-10T10:04:04

$tan ({sin}^{- 1} (- \frac{4}{5})) = \pm \frac{4}{3}$ $tan(sin^(-1)(-4/5))=+-4/3$

Let $sin θ = (- \frac{4}{5})$ $sintheta=(-4/5)$ ............(A)
where $θ = {sin}^{- 1} (- \frac{4}{5})$ $theta=sin^(-1)(-4/5)$

Hence as ${cos}^{2} θ = 1 - {sin}^{2} θ$ $cos^2theta=1-sin^2theta$ , we have

$cos θ = \sqrt{1 - {sin}^{2} θ} = \sqrt{1 - {(- \frac{4}{5})}^{2}}$ $costheta=sqrt(1-sin^2theta)=sqrt(1-(-4/5)^2$ or

$cos θ = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \pm \frac{3}{5}$ $costheta=sqrt(1-16/25)=sqrt(9/25)=+-3/5$ ............(B)

Dividing (A) by(B), we get

$tan θ = \pm \frac{4}{3}$ $tantheta=+-4/3$ or

$tan ({sin}^{- 1} (- \frac{4}{5})) = \pm \frac{4}{3}$ $tan(sin^(-1)(-4/5))=+-4/3$

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