The formula is proved tautology using Truth Table but gives another answer if solved with formula I am trying to prove by formula that the following formula is a tautology but I am not able to do so, although I have proved it as a tautology with truth table  
 (A∨B) ∧ (¬B∨C) → A∨C

¬((A∨B) ∧ (¬B∨C)) ∨  A∨C
 (¬A∧¬B) ∨ (B∧¬C)  ∨  A∨C
¬A∨(B∧¬C) ∧ ¬B∨(B∧¬C) ∨  A∨C              (Distributive Law)
(¬A∨B) ∧ (¬A∨¬C) ∧ (¬B∨B) ∧ (¬B∨¬C ∨ A∨C )  
(¬A∨B) ∧ (¬A∨¬C) ∧  T   ∧  (¬B∨A ∨  T) 
∴(¬A∨B) ∧ (¬A∨¬C) 

By using Truth table, I am able to prove it as a tautology. 
Am I doing something wrong in the second method?
 A: Your mistake is here:
¬A∨(B∧¬C) ∧ ¬B∨(B∧¬C) ∨  A∨C              (Distributive Law)
(¬A∨B) ∧ (¬A∨¬C) ∧ (¬B∨B) ∧ (¬B∨¬C ∨ A∨C )  

The result should be:
¬A∨(B∧¬C) ∧ ¬B∨(B∧¬C) ∨  A∨C              (Distributive Law)
((¬A∨B) ∧ (¬A∨¬C) ∧ (¬B∨B) ∧ (¬B∨¬C)) ∨ A∨C   

So be careful with those parentheses! In fact, I believe your mistake was partly based on the fact that you didn't use parentheses in the statement before. That is, instead of:
¬A∨(B∧¬C) ∧ ¬B∨(B∧¬C) ∨  A∨C              (Distributive Law)

I would use parentheses to make the structure of the statement really clear:
((¬A∨(B∧¬C)) ∧ (¬B∨(B∧¬C))) ∨  A∨C              (Distributive Law)

Doing that makes it less likely to make mistakes.
By the way, to continue to put this in to CNF (you asked about this in the comments):
((¬A∨B) ∧ (¬A∨¬C) ∧ (¬B∨B) ∧ (¬B∨¬C)) ∨ A∨C (Complement)
((¬A∨B) ∧ (¬A∨¬C) ∧ $\top$ ∧ (¬B∨¬C)) ∨ A∨C (Identity)
((¬A∨B) ∧ (¬A∨¬C) ∧ (¬B∨¬C)) ∨ A∨C (Distribution)
((¬A∨B) ∨ (A∨C)) ∧ ((¬A∨¬C) ∨ (A∨C)) ∧ ((¬B∨¬C) ∨ (A∨C)) (Association)
(¬A∨B ∨ A∨C) ∧ (¬A∨¬C ∨ A∨C) ∧ (¬B∨¬C ∨ A∨C) (Complement x 3)
($\top$ ∨B ∨C) ∧ ($\top$ ∨¬C ∨C) ∧ ($\top$ ∨¬B∨ A) (Identity x 3)
$\top$  ∧ $\top$  ∧ $\top$  
$\top$
(you're not going to get any 'interesting' clauses exactly because this is a tautology!)
Also, here is an easier way to show that your statement is a tautology. It relies on a little known equivalence called Concensus:
$(P \lor Q) \land (\neg Q \lor R) \Leftrightarrow (P \lor Q) \land (\neg Q \lor R) \land (P \land R)$
(It's basically the Resolution rule, but as an equivalence)
Applied to your statement:
$((A \lor B) \land (\neg B \lor C)) \rightarrow (A \lor C) \Leftrightarrow$ (Concensus)
$((A \lor B) \land (\neg B \lor C) \land (A \lor C)) \rightarrow (A \lor C) \Leftrightarrow $ (Implication)
$\neg ((A \lor B) \land (\neg B \lor C) \land (A \lor C)) \lor (A \lor C) \Leftrightarrow$ (DeMorgan)
$\neg (A \lor B) \lor \neg (\neg B \lor C) \lor \neg (A \lor C) \lor (A \lor C) \Leftrightarrow$ (Complement)
$\neg (A \lor B) \lor \neg (\neg B \lor C) \lor \top \Leftrightarrow$ (Identity)
$\top$
A: It looks like you may have forgotten some parentheses here. As a rule of thumb parentheses (even when slightly overused) can make a huge difference in Logic. For example in your approach: 
$$¬[(A∨B) ∧ (¬B∨C)] ∨  (A∨C)$$
$$ [(¬A∧¬B) ∨ (B∧¬C)]  ∨  (A∨C)$$
$$[¬A∨(B∧¬C) ∧ ¬B∨(B∧¬C)] ∨  (A∨C)$$           

$$[(¬A∨B) ∧ (¬A∨¬C) ∧ (¬B∨B) ∧ (¬B∨¬C)] ∨ (A∨C)$$

And this is where the mistake happened.
However, there is an easier approach from step 2 above. Incidentally you can drop/re-arrange, a lot of parentheses because the clauses share the same logical connective. Namely: 
$$¬[(A∨B) ∧ (¬B∨C)] ∨  (A∨C)$$
$$ [(¬A∧¬B) ∨ (B∧¬C)]  ∨  (A∨C)$$
$$ (¬A∧¬B) ∨ (B∧¬C)  ∨  (A) ∨ (C)$$
$$ (¬A∧¬B) ∨ (A) ∨ (B∧¬C)  ∨ (C)$$
$$ [(¬A∧¬B) ∨ (A)] ∨ [(B∧¬C)  ∨ (C)]$$
$$ [(¬A∨A)∧(¬B∨A)] ∨ [(B∨C) ∧(¬C ∨ C)]$$
$$ [T∧(¬B∨A)] ∨ [(B∨C) ∧ T]$$
$$ (¬B∨A) ∨ (B∨C)$$
$$ (¬B)∨ (A) ∨ (B) ∨(C) $$ 
$$ (¬B)∨ (B) ∨ (A) ∨(C) $$ 
$$ (¬B∨B) ∨ (A) ∨(C) $$ 
$$ T ∨ (A) ∨(C) $$
$$T$$ 
