How do you find $tan ({sin}^{- 1} (\frac{p}{\sqrt{p^{2} + 9}}))$ $tan(sin^-1(p/sqrt(p^2+9)))$ ?

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Answer 1 · 2016-09-01T04:51:56

$\frac{p}{3}$ $p/3$

Let $a = {sin}^{- 1} (\frac{p}{\sqrt{p^{2} + 9}})$ $a = sin^(-1)(p/sqrt(p^2+9))$

$in Q1 if p >0, and Q4, if p < 0.$ , for the principal valur.

In both, cos a is positive.

Then, $sin a = p/sqrt(p^2+9)$

$cos a = sqrt(1-p^2/((p^2+9))$

$=3/sqrt(p^2+9)$ and the given expression

$tana = sin a/cos a= p/3$ .

If general values are considered, a could be in any of the four

quadrants and tan a = $+-p/3$ .
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How do you find tan(sin^-1(p/sqrt(p^2+9)))tan(sin−1(p√p2+9))tan(sin^-1(p/sqrt(p^2+9)))?