Answer
$A_{0}(x)=1\\A_{1}(x)=1+2(x-\dfrac{\pi}{4}) \\A_{2}(x)=1+2(x-\dfrac{\pi}{4})+(x-\dfrac{\pi}{4})^2 \\A_{3}(x)=1+2(x-\dfrac{\pi}{4})+(x-\dfrac{\pi}{4})^2+\dfrac{8}{3}(x-\dfrac{\pi}{4})^3$
Work Step by Step
The Taylor polynomial is defined for the function $f(x)$ at the point $a$ of order $n$ as:
$A_n(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$
Now, we get $f(\dfrac{\pi}{4})=1 \\f'(x)=sec^2 x \implies f'(\dfrac{\pi}{4})=2 \\ f''(x)=2 \sec^2 x \tan x \implies f''(\dfrac{\pi}{4})=-4 \\ f'''(x)=4 \sec^2 x \tan^2 x+2 \sec^4 x \implies f'''(\dfrac{\pi}{4})=16$
This implies that $A_{0}(x)=1\\A_{1}(x)=1+2(x-\dfrac{\pi}{4}) \\A_{2}(x)=1+2(x-\dfrac{\pi}{4})+(x-\dfrac{\pi}{4})^2 \\A_{3}(x)=1+2(x-\dfrac{\pi}{4})+(x-\dfrac{\pi}{4})^2 +(\dfrac{16}{6})(x-\dfrac{\pi}{4})^3=1+2(x-\dfrac{\pi}{4})+(x-\dfrac{\pi}{4})^2+\dfrac{8}{3}(x-\dfrac{\pi}{4})^3$
