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#### Four Terms in The Expansion

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## Find the first four terms in the expansion of log(1+tanx) in the power of x.

Let,
f(x) = log (1+tan x)
or f’(x) =1/(1+tan x) *sec2 x
or f”(x)= d/dx (sec2 x/ 1 + tan x)
or f”(x)=[(1+tan x) (2sec x. sec x. tan x) – sec2 x (sec2 x)] / (1 + tan x)2
Now, f(0) = log(1+tan 0 ) = log (11=0)
F’(0)= 1/1+0=1
Or f”(0)= -1
Therefore,
f”(x)= [(1 + tan x) (2sec2 x * tan x) – sec2 x] / (1+tan x)2
or f”(x)= [2sec2 x. tan x + 2sec4 x – 2 sec2 x – sec4 x] / (1+ tan x)2
or f”(x)= [(2 sec2 x (tan x -1) + sec 4 x)] / (1+tan x)2
or f”(x)= (1 + tan x)2 [4 sec x. sec x. tan x (tan x -1) + 2 sec2 x (sec2 x) + 4 sec3 x. sec x. tan x] – [2 sec2 x (tan x + 1) + sec4 x]
Solving for f”(x) ; x = 0
f(x) = f(0) + f’(0). x + f”(0)/ 2!b* x2 + f”’ (0)/ 3! x3   + f “’’ (0)/ 4! * x4 + …. Infinity
(Ans) or f(x) = 0 + x + (-1x2 /2!) + (-6/3!)* x3 + f “’’ (0)/ 4! * x4

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