Trigonometry

Usa a right triangle to write each expression as an algebraic expression. Assume the referenced angle is theta, and x is positive and in the domain of the given inverse trig. function. Label the triangle.

1. sec^sq(tan^-1x)

2. sin(arccot x/(sqrt)1-x^2)

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  1. 1. If you mean sec^2[tan^-1x], use the fact that
    sec (tan^-1x)= sqrt(x^2+1)/sqrt(1).
    Imagine a right triangle with sides x, 1 and sqrt(1+x^2). Side x is opposite the angle with tangent equal to x.
    Square the secant and you get x^2 + 1

    2. For the angle in question, as part of a right triangle:
    adjacent side (not hypotentuse) = x
    opposite side = sqrt(1-x^2)
    hypotenuse = 1
    sin = sqrt(1-x^2)
    The last line is your answer.

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    posted by drwls

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