I need help with a math question involving exact values

These are the questions:

(a) tan(csc^-1(2))
(b) sin(2cos^-1(3/5))

Can someone explain how to find the exact value of each of these expressions?

csc⁻¹ always returns a value between -π/2 and π/2

Let θ = csc⁻¹(2)
csc(θ) = 2
sin(θ) = 1/2
θ = π/6

tan(csc⁻¹(2)) = tan(θ) = tan(π/6) = 1/√3

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cos⁻¹ always returns a value between 0 and π

Let θ = cos⁻¹(3/5)
cos(θ) = 3/5

sin²(θ) = 1 - cos²(θ)
sin²(θ) = 1 - 9/25
sin²(θ) = 16/25

cos⁻¹ always returns value between 0 and π
---> θ is in quadrant I or II
---> sin(θ) > 0, so we take positive square root

sin(θ) = 4/5

sin(2cos⁻¹(3/5)) = sin(2θ)
. . . . . . . . . . . . . = 2 sin(θ) cos(θ)
. . . . . . . . . . . . . = 2 * 4/5 * 3/5
. . . . . . . . . . . . . = 24/25

-- Ματπmφm --

Mike's answer is incorrect (though more satisfying, admittedly) since he doesn't restrict domains.

Notes: r(x) is the positive square root of x.

a. csc^-1(2) = sin^-1(1/2). cos(sin^-1(1/2)) = r(3)/2 > 0 by domain restriction. Then tan(csc^-1(2)) = 1/2/(r(3)/2) = r(3)/3

b. sin(2a) = 2sin(a)cos(a). cos(cos^-1(3/5)) = 3/5. sin(cos^-1(3/5))>0 by domain restriction. Then sin(") = 4/5. Hence, sin(2cos^-1(3/5)) = 2*3*4/25 = 24/25

(a) tan(csc⁻ֿ¹(2))

csc⁻ֿ¹(2) = A
csc A = 2
1/sinA = 2
sinA = 1/2
A = sin⁻ֿ¹(1/2)

tan(csc⁻ֿ¹(2)) = tan(sin⁻ֿ¹(1/2))
= sin(sin⁻ֿ¹(1/2)) / cos(sin⁻ֿ¹(1/2))
= [1/2] / √ (1 - sin²(sin⁻ֿ¹(1/2))
= ±[1/2] / √ (1 - 1/4)
= ±[1/2] / [√ (3)/2]
= ±1/√ 3

(b) sin(2cos⁻ֿ¹(3/5))
note sin(2A) = 2sinAcosA
sin(2cos⁻ֿ¹(3/5)) = 2sin(cos⁻ֿ¹(3/5)) cos(cos⁻ֿ¹(3/5))
= 2 √[1 - cos²(cos⁻ֿ¹(3/5) ] cos(cos⁻ֿ¹(3/5))
= 2 √[1 - (3/5)²] (3/5)
= 6√[16/25] /5
= 6(4/5) /5
= 24/25

a) tan(csc^-1(2))
csc A = 2
sin A = 1/2
A =- 30degrees.............domain of Inverse sin and csc funtions

tan (-30degrees)=-sqrt(3)/3

b) sin(2cos^-1(3/5))

Draw a right triangle in the first quadrant with horizontal leg 3 and hypotenuse 5
The vertical leg then must be 4
Call the angle A

sin(2A)

sin(2A)=2sinAcosA

sin(2A) = (2/1)*(4/5)*(3/5)

ANSWER...........24/25

(a) tan(csc^-1(2))
cosec A = 2
sin A = 1/2
A = 30 or 150 degrees
tan A = 1/sqrt(3) or -1/sgrt(3)
tan A = sqrt(3)/3 or -sqrt(3)/3