Find the value of $tan ({cos}^{- 1} (\frac{3}{x}))$ $tan(cos^(-1)(3/x))$ ?

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Answer 1 · 2017-10-24T17:11:24

$tan ({cos}^{- 1} (\frac{3}{x})) = tan t = \frac{1}{3} \sqrt{x^{2} - 9}$ $tan(cos^(-1)(3/x))=tant=1/3sqrt(x^2-9)$

Let ${cos}^{- 1} (\frac{3}{x}) = t$ $cos^(-1)(3/x)=t$

then $cos t = \frac{3}{x}$ $cost=3/x$

and therefore $sec t = \frac{x}{3}$ $sect=x/3$

and ${sec}^{2} x = \frac{x^{2}}{9}$ $sec^2x=x^2/9$ and ${tan}^{2} t = \frac{x^{2}}{9} - 1 = \frac{x^{2} - 9}{9}$ $tan^2t=x^2/9-1=(x^2-9)/9$

and $tan t = \frac{1}{3} \sqrt{x^{2} - 9}$ $tant=1/3sqrt(x^2-9)$

hence $tan ({cos}^{- 1} (\frac{3}{x})) = tan t = \frac{1}{3} \sqrt{x^{2} - 9}$ $tan(cos^(-1)(3/x))=tant=1/3sqrt(x^2-9)$

Find the value of tan(cos^(-1)(3/x))tan(cos−1(3x))tan(cos^(-1)(3/x))?