what is the exact value of tan(arctan3/4-arcsin1/2)?

Answer 1

a = arctan 3/4
tan a = 3/4

b = arcsin 1/2
sin b = 1/2
tan b = 1/sqrt(3)

tan (arctan 3/4-arcsin 1/2)
= tan (a - b)
= (tan a - tan b) / (1 + tan a tan b)
= (3/4 - 1/sqrt(3)) / (1+ (3/4)(1/sqrt(3)))
= (3sqrt(3) - 4) / (4sqrt(3) + 3)

Answer 2

arcsin(1/2) is the solution to sin(x) = 1/2 but with the restriction
-pi/2 <= x <= pi/2. Therefore, arcsin(1/2) = pi/6.

Also, arctan and tan are inverses of each other, so
tan(arctan(3/4)) = 3/4.
Using the identity

tan(x - y) = [ tan(x) - tan(y) ] / [1 + tan(x)tan(y)]

tan(arctan(3/4) - pi/6) =
[tan(arctan(3/4) - tan(pi/6)] / [1 + tan(arctan(3/4))tan(pi/6)]

= [3/4 - 1/sqrt(3)] / [1 + (3/4)(1/sqrt(3))]

Multiply top and bottom by 4sqrt(3) to eliminate the complex fraction,

= [3sqrt(3) - 4] / [sqrt(3) + 3]

Now, we rationalize the denominator by multiplying top and bottom by the bottom's conjugate. The bottom becomes a difference of squares.

= [3sqrt(3) - 4] [sqrt(3) - 3] / [ 3 - 9 ]
= [3(3) - 9sqrt(3) - 4sqrt(3) + 12] / [-6]
= [9 - 13sqrt(3) + 12] / (-6)
= [21 - 13sqrt(3)] / (-6)
= (-1/6)(21 - 13sqrt(3))