Hi I am not sure I am solving this trig question correctly:
$\tan\left(\sin^{-1}\left(\dfrac{2\sqrt{x}}{1+x}\right)\right) = ?$
I drew a right triangle and set an angle equal to $\sin^{-1}\left(\dfrac{2\sqrt{x}}{1+x}\right)$
so that I could find the third (adjacent) side after setting the opposite as $2\sqrt{x}$ and the hypotenuse as $1+x$ by using the Pythagorean theorem.
I believe after using the Pythagorean theorem and doing the algebra that the adjacent side is $\sqrt{x^2-2x+1}$
So would it then follow that the answer to the question would be $\dfrac{2\sqrt{x}}{\sqrt{x^2-2x+1}}$?
thanks