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tan(2sin^-1x). Write the expression as an algebraic expression in x (without trig or inverse trig functions)
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\[\text {Let} \theta = \sin^{-1} x\] \[\sin \theta = x \] now goal is to find \[\tan 2 \theta\] \[\rightarrow \tan 2 \theta = \frac{\sin 2\theta}{\cos 2 \theta} = \frac{2 \sin \theta \cos \theta}{\cos^2 \theta - \sin^2 \theta}\] Find cos using pythagorean identity \[\cos \theta = \sqrt{1-x^2}\] plug in the values and you are done
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