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^(-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.

Integral = [x^3 + 2x] from 0 to 1 = (1 + 2) - (0) = 3.

The integral evaluates to [x^3] from 0 to 1 = 1^3 - 0^3 = 1.

The integral ∫(0 to 1) x^2 dx = [x^3/3] from 0 to 1 = 1/3.

The integral evaluates to [x^3 - 2x] from 1 to 2 = (8 - 4) - (1 - 2) = 3.

The integral evaluates to 7.

The integral evaluates to tan^-1(x) + C.

∫(2x)dx = x^2 from 0 to 2 = 2^2 - 0^2 = 4.

The integral evaluates to 20.

Evaluating the integral gives [x^3 + x^2]_0^1 = (1 + 1) - (0 + 0) = 2.

Using L'Hôpital's Rule, the limit evaluates to 2.

The limit lim(x->0) (sin(x)/x) = 1.