Using the co-function identity, sin(90° - x) = cos(x).

sin(90°) = 1.

sin(tan^(-1)(1)) = √2/2

sin(tan^(-1)(x)) = x/√(1+x^2)

sin(tan^(-1)(√3)) = √3/2

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

sin^(-1)(1/2) = π/6 and sin^(-1)(√3/2) = π/3. Therefore, π/6 + π/3 = π/2.

sin^(-1)(1/2) = π/6, since sin(π/6) = 1/2.

sin^(-1)(sin(π/4)) = π/4, as π/4 is in the range of sin^(-1).

By the Pythagorean identity, sin^2(x) + cos^2(x) = 1 for all x.

By the Pythagorean identity, sin^2(θ) + cos^2(θ) = 1.

tan(30°) = sin(30°)/cos(30°) = (1/2)/(√3/2) = 1/√3.

Using the formula tan(A + B) = (tan A + tan B) / (1 - tan A tan B), we have tan(45° + x) = (1 + tan x) / (1 - 1*tan x) = (1 + tan x).

tan(45°) = 1.

tan^(-1)(1) = π/4, thus tan^(-1)(1) + tan^(-1)(1) = π/4 + π/4 = π/2.

Using the formula tan^(-1)(a) + tan^(-1)(b) = tan^(-1)((a+b)/(1-ab)), we have tan^(-1)(1) + tan^(-1)(2) = tan^(-1)((1+2)/(1-1*2)) = tan^(-1)(3/-1) = π - tan^(-1)(3) = π/4.

tan^(-1)(1) = π/4, since tan(π/4) = 1.

tan^(-1)(√3) corresponds to the angle π/3.

The 5th term is given by C(6,4) * (x)^4 * (2)^2 = 15 * x^4 * 4 = 240.

The coefficient of x^3 is C(6,3) * (-1)^3 = 20 * (-1) = -20.

The coefficient of x^5 is C(7,5) * (3)^2 = 21 * 9 = 189.

The coefficient of x^5 is C(8, 5) * (3)^3 = 56 * 27 = 1512.

f'(x) = (2x)/(x^2 + 1). At x = 1, f'(1) = 2/(1 + 1) = 1.

The determinant is calculated as (1*4) - (2*3) = 4 - 6 = -2.

The determinant of matrix A is 0 because the rows are linearly dependent.

The determinant of the matrix is 0 because the rows are linearly dependent.

The determinant is 0 because the first column is repeated.

The determinant is \( 1*4 - 2*3 = 4 - 6 = -2 \).

Calculating the determinant gives -12.

det = (2)(5) - (3)(4) = 10 - 12 = -2.