Find the Taylor Series expansion of tan(x) about π/4 up to the term in terms of (x-π/4)^3.

Firstly, set f(x) = tan(x). We can then differentiate this 3 times in order to find f'(x), f''(x) and f'''(x). This will require use of the chain rule and the product rule. We can find:f'(x) = sec²xf''(x) = 2 * sec(x) * sec(x)tan(x) = 2sec²(x)tan(x)f'''(x) = 2 * 2sec²(x)tan(x)tan(x) + 2sec²(x)*sec²(x) = 4sec²(x)tan²(x) + 2sec⁴(x)We then substitute x with π/4, and find the values of f at these values, and then the coefficient a_n:f(π/4) = 1 a₀ = 1/0! = 1 f'(π/4) = 2 a₁ = 2/1! = 2f''(π/4) = 4 a₂ = 4/2! = 2f'''(π/4) = 16 a₃ = 16/3! = 8/3From this we can conclude tan(x) = 1 + 2(x-π/4) + 2(x-π/4)² + (8/3)(x-π/4)^3.