cos^(-1)(0) = π, since cos(π) = 0.

sin^(-1)(sin(π/4)) = π/4

If sin(x) = 1/√2, then x = π/4, thus tan(sin^(-1)(1/√2)) = tan(π/4) = 1.

tan^(-1)(√3) = π/3, since tan(π/3) = √3.

sin^(-1)(x) + cos^(-1)(x) = π/2 for all x in the domain [-1, 1].

Using the identity sin^(-1)(x) + sin^(-1)(√(1-x^2)) = π/2 for x in [0, 1], the value is π/2.

2sin^(-1)(1/2) = 2(π/6) = π/3 and 2cos^(-1)(1/2) = 2(π/3) = 2π/3. Therefore, the total is π/3 + 2π/3 = π.

By definition, \( \cos(\cos^{-1}(x)) = x \). Therefore, \( \cos(\cos^{-1}(\frac{3}{5})) = \frac{3}{5} \).

Using the identity, cos(tan^(-1)(3/4)) = 4/5

cos^(-1)(0) = π/2, since cos(π/2) = 0.

sin^(-1)(√3/2) = π/3 and cos^(-1)(1/2) = π/3. Therefore, the sum is π/3 + π/3 = 2π/3.

By definition, \( \sin(\sin^{-1}(x)) = x \). Therefore, \( \sin(\sin^{-1}(\frac{1}{2})) = \frac{1}{2} \).

x = cos^(-1)(-1/2) = 2π/3

If x = cos^(-1)(1/2), then x = π/3. Therefore, sin(x) = sin(π/3) = √3/2.

If x = cos^(-1)(1/2), then cos(x) = 1/2, which corresponds to x = π/3. Therefore, sin(x) = sin(π/3) = √3/2.

If x = cos^(-1)(1/2), then x = π/3, thus sin(x) = sin(π/3) = √3/2.

sin^(-1)(-1) corresponds to the angle -π/2.

sin^(-1)(-1) corresponds to the angle -π/2.

If x = sin^(-1)(1/2), then x = π/6. Therefore, cos(x) = cos(π/6) = √3/2.

Using the identity cos(x) = √(1 - sin^2(x)), we find cos(sin^(-1)(1/3)) = √(1 - (1/3)^2) = √(8)/3.

tan^(-1)(1) = π/4.

tan^(-1)(1/√3) = π/6, since tan(π/6) = 1/√3.

The derivative of cos^(-1)(x) is dy/dx = -1/√(1-x^2).

The derivative of y = sin^(-1)(x) is dy/dx = 1/√(1-x^2).

By definition, if y = sin^(-1)(x), then x = sin(y).

The second derivative d^2y/dx^2 = -1/√(1-x^2)^3.

The range of y = tan^(-1)(x) is (-π/2, π/2).

The first derivative dy/dx = 1/(1+x^2). The second derivative d^2y/dx^2 = -2/(1+x^2)^2.

The derivative of \( y = \cot^{-1}(x) \) is \( -\frac{1}{1+x^2} \).

The derivative of \( y = \sec^{-1}(x) \) is \( \frac{1}{|x|\sqrt{x^2-1}} \).