Show that? : $tan(arcsinx) = x/sqrt(1 -x^2)$

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Answer 1 · 2017-05-02T02:16:37

Let

$y=arcsinx iff x=siny$

Then using $sin^2A+cos^2A -= 1$ ; we have:

$sin^2y+cos^2y = 1 => x^2+cos^2y=1$
$:. cos^2y = 1 -x^2$
$:. sec^2y = 1/(1 -x^2)$

And, using the trig identity $tan^2A+1-=sec^2A$ ; we have:

$tan^2y+1-=sec^2y$

$:. tan^2y = sec^2y - 1$
$" " = 1/(1 -x^2) - 1$
$" " = (1-(1-x^2))/(1 -x^2)$
$" " = (x^2)/(1 -x^2)$

And so:

$:. tany = sqrt((x^2)/(1 -x^2))$
$" " = x/sqrt(1 -x^2)$

But $y=arcsinx$ ; therefore:

$tan(arcsinx) = x/sqrt(1 -x^2) \ \ \$ QED

Show that? : tan(arcsinx) = x/sqrt(1 -x^2) tan(arcsinx) = x/sqrt(1 -x^2)