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)
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
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)
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)
1 answer