Approximate tan(-.1) using Taylor polynomial with n=3

I had this one but lost how to do it in my notes...
I know that we use f(0) f'(0) f''(0) and f'''(0)
Since prime and triple prime have nonzero values do we take their functions, add and evaluate at -.1? The answer should be .100333... But I can't seem to get that

f(x) = tan x ==> f(0) = 0

f '(x) = sec^2(x) ==> f '(0) = 1

f ''(x) = 2 sec^1(x) * sec x tan x
.......= 2 sec^2(x) tan x ==> f ''(0) = 0

f '''(x) = 4 sec^1(x) sec x tan x * tan x + 2 sec^2(x) * sec^2(x) ==> f '''(0) = 2.

Hence, tan x = 0 + 1x + 0x^2/2! + 2x^3/3! + ...
....................= x + x^3/3 + ...

So, tan(-0.1) ≈ -0.1 + (-0.1)^3/3 = -0.100333...
(Note: This answer should be negative.)

I hope this helps!