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How do I write tan(2 sin^-1(x)) as an algebraic expression?

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  1. Consider a right-angled triangle with hypotenuse = 1, and opposite = x

    sinθ = opposite/hypotenuse = x/1
    then by Pythagoras the adjacent side = √(1-x^2)
    and cosθ = √(1-x^2)

    sin^-1(x) is the angle θ so that sinθ = x
    then tan(2 sin^-1 (x) )
    = tan (2θ)
    = sin 2θ / cos 2θ
    = 2sinθcosθ/(cos^2 θ - sin^2 θ)
    = 2x√(1-x^2) / (1-x^2 - x^2)
    = 2x√(1-x^2) / (1- 2x^2)

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