Does conditional tautology mean a tautology which hast the form of $(A)\rightarrow (b)$ and therefore $A \lor \lnot A$ is unconditional tautology?
(in regards to the following paragraph from wikipedia about sequent calculus: "https://en.wikipedia.org/wiki/Sequent_calculus")
Sequent calculus is, in essence, a style of formal logical argumentation where every line of a proof is a conditional tautology (called a sequent by Gerhard Gentzen) instead of an unconditional tautology.
In sequent calculus, you have a sequent, often written $\Gamma \vdash P$ where $\Gamma$ is a list of propositions, and $P$ is a proposition (or $P$ might also be a list of propositions in which case the metavariable $\Delta$ is often used producing $\Gamma\vdash\Delta$). Typically (but sometimes the notation is used in slightly different ways), this means given the propositions in $\Gamma$ hold, $P$ holds. So an unconditional tautology using this notation looks like $\vdash P$.
Unsurprisingly, the rules for $\to$ essentially internalize this meta-theoretic conditional reasoning:$$\cfrac{\Gamma,P\vdash Q}{\Gamma\vdash P\to Q}$$ So, given implication, every conditional tautology, $P\vdash Q$, can be reduced to an unconditional tautology: $\vdash P\to Q$. Just do it repeatedly if there is more than one assumption: $P,Q\vdash R \iff \vdash P\to(Q\to R)$, or you can use conjunction to combine them if you have conjunction: $P,Q\vdash R \iff \vdash(P\land Q)\to R$.
A logic need not have implication, i.e. $\to$, or any equivalent, so it's not necessarily the case that we can always reduce a conditional tautology to an unconditional one. Even when a logic does have $\to$, it is often more natural to be able to directly talk about hypothetical reasoning. Obviously this is useful for implication, but also for negation and to some extent proof-by-cases, i.e. disjunction elimination. In that last case, if you want to use an assumption of the form $P\lor Q$, e.g. to prove $(P\lor Q)\to R$, the reasoning is often structured as "if $P$ holds, I can prove $R$, and if $Q$ holds, I can prove $R$, therefore if either $P$ or $Q$ holds, I can prove $R$". Indeed, this is what the natural deduction form of disjunction elimination states: $$\cfrac{\Gamma\vdash P\lor Q\qquad\Gamma, P\vdash R\qquad \Gamma, Q\vdash R}{\Gamma\vdash R}$$ Using only unconditional tautologies would require something like: $$\cfrac{\vdash P\lor Q\qquad\vdash P\to R\qquad \vdash Q \to R}{\vdash R}$$ This involves $\to$ in the rules for $\lor$, so with this presentation we can't talk about $\lor$ without also understanding $\to$. It gets worse. We've lost $\Gamma$, so the previous rule is, on its own, weaker than the earlier version. We can fix this as follows:$$\cfrac{\vdash S\to(P\lor Q)\qquad\vdash S\to(P\to R)\qquad \vdash S\to(Q \to R)}{\vdash S\to R}$$ But $\Gamma$ is a list of propositions, not a single propositions, so this is still weaker than the first rule unless we have conjunction. So now to fully understand $\lor$ we need to understand $\to$ and $\land$. This tangles together the meaning of the logical connectives. With the first presentation of disjunction elimination, we can give meaning to $\lor$ in logics which don't have $\to$ or $\land$.
(As an aside, you can start to see how the Hilbert axioms arise and why they are the way they are. For example, the S axiom: $(P\to(Q\to R))\to((P\to Q)\to (P\to R))$ is closely related to $\to$ elimination (i.e. modus ponens): $$\cfrac{\Gamma\vdash Q\to R\qquad \Gamma\vdash Q}{\Gamma\vdash R}$$ We see the axiom is the same kind of unconditional rendition of the rule where $P$ stands for the context $\Gamma$.)
