Chapter 10: Infinite Sequences and Series - Section 10.8 - Taylor and Maclaurin Series - Exercises 10.8 - Page 619: 41

$L(x)=0$ and $Q(x)=\dfrac{-x^2}{2!}$

Differentiate the given function $f(x)$ as follows: $f'(x)=-\tan x ; \\f''(x)=-\sec^2 x $ Now $ f(0)=0; \\ f'(0)=0; \\ f'(0)=-1$ Therefore, $L(x)=f(0)+xf'(0)=0$ and $Q(x) =f(0) x +xf'(0) +\dfrac{x^2 f''(0)}{2!}=\dfrac{-x^2}{2!}$ Our result is: $L(x)=0$ and $Q(x)=\dfrac{-x^2}{2!}$