1. Evaluate tan[sin^-1(a)].

a. [sqrt(1-a^2)]/1-a^2
b. [sqrt(1-a^2)]/a
c. sqrt(1-a^2)
d. {a[sqrt(1-a^2)]}/1-a^2

I do not know the steps to find this answer and am not provided with a textbook. I have researched online some what, but I cannot seem to understand. Can someone please explain this?

3 answers

draw the triangle. You want the angle whose sine is a = a/1
so, pick an acute angle, and its opposite side is a, and the hypotenuse is 1.
That means the adjacent side is √(1-a^2)

so, tan(arcsin(a)) = opposite/adjacent = a/√(1-a^2)

which, for those who insist on rational denominators, is (d)
Oh my goodness. This makes so much more sense than the extremely confusing lesson this is in. Thank you Steve!
right triangle sides a b c and angles A B C
C is right, c is hypotenuse
sin A = a/c
let c = 1
sin A = a
a^2 + b^2 = 1 so b = sqrt(1-a^2)
tan A = a/b = a/sqrt(1-a^2)
multiply top and bottom by sqrt(1-a^2) to get rid of sqrt in denominator
= a sqrt(1-a^2) / (1-a^2)
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