Answer
\[\boxed{\begin{array}{*{20}{c}}
n&{{x_n}}&{f\left( {{x_n}} \right)}&{f'\left( {{x_n}} \right)}&{\frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}}&{{x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}} \\
1&{1.6000}&{0.100334}&{1.01006}&{0.099334}&{0.000665995} \\
2&{0.000665}&{0.000665}&{1.00000}&{0.000665}&{0.000005985}
\end{array}}\]
Work Step by Step
$$\eqalign{
& f\left( x \right) = \tan x,{\text{ }}{x_1} = 0.1 \cr
& {\text{Use Newton's Method}} \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr
& f\left( {{x_1}} \right) = \tan \left( {0.1} \right) \approx 0.100334 \cr
& {\text{Use Newton's Method}} \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}{\text{ }}\left( {\bf{1}} \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\tan x} \right] = {\sec ^2}x \cr
& f'\left( {{x_1}} \right) = {\sec ^2}\left( {0.1} \right) \approx 1.010067 \cr
& {\text{Substituting }}f\left( x \right){\text{ and }}f'\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {x_{1 + 1}} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}} \cr
& {x_2} = 0.1 - \frac{{0.100334}}{{1.010067}} \approx 0.000665995 \cr
& \tan \left( {0.000665995} \right) \approx 0.000665995 \cr
& {\text{The calculations for two iterations are shown in the table}} \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
n&{{x_n}}&{f\left( {{x_n}} \right)}&{f'\left( {{x_n}} \right)}&{\frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}}&{{x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}} \\
1&{1.6000}&{0.100334}&{1.01006}&{0.099334}&{0.000665995} \\
2&{0.000665}&{0.000665}&{1.00000}&{0.000665}&{0.000005985}
\end{array}}\]
