We increase voltage by using step up transformer so that current reduces and voltage increases and consequently energy losses (generally heat) are reduced.
$$E=I^2 Rt$$ But energy is also given as: $$E=VIt$$ Both formulae are exactly equivalent. Now question is that energy depends on both V & I. then how does step up transformer reduces energy loss by increasing V and decreasing I?
I'm not sure what you mean by "exactly equivalent". In particular, the $E$ in your equations indicates different things.
In the second equation, $E$ is the total energy delivered by the circuit over the period of time (the load on the system). In a power delivery system, this will be driven by customer demand.
In the first equation, $E$ is the energy loss associated with a fixed-resistance portion. Often used when looking at the transport wires for instance. The trick of reducing resistive losses only works when they account for a portion of the power delivered, not the entire amount.
Let's imagine that instead of the circuit being dominated by a somewhat-constant customer power load, all of the power goes into a single resistor element $R$.
Assume we begin with $1V$ driving $1A$ through $1 \Omega$. The power delivered is $1W$.
Now to lower the resistance loss, we try to keep our power constant, but raise the voltage and lower the current. Unfortunately, without the large constant load, this is impossible. By raising the voltage to $2V$, the constant resistance now allows $2A$ to flow. Instead of reducing the power loss or holding the power delivered constant, both were raised.
Transformers use AC currents as they work on the principle of mutual induction. Therefore, we should use the RMS values of voltage and current for understanding power.
What you have failed to notice is that the net power ouput and the power loss due to heat is different. For the sake if simplicity, let us consider resistance to be the only source of energy loss.
The rate of energy loss due to the resistance is given by: $$P = I^2 R$$
As you see, the resistance of the wire remains constant. Therefore, we can conclude that energy loss increases squarely with the increase in current.
Therefore, larger current would mean more energy loss.
You have to be careful when you draw conclusions from the following formula:
$$P = VI$$
The symbol $V$ here represents the voltage across the resistance, i.e: wire. The total voltage across tbe terminals is different.
Thetefore, when we say we increase voltage and reduce current, we mean to say we increase the voltage across the terminals of the transformer. We do not increase the voltage across the wire. Thid is where you confusion lies.
The following is a simple model of a cable and a load:
The power delivered to the load is given by
$P_L=I^2R_L$
The power wasted by the transmission cable is:
$P_T=I^2R_T$
The current going through the circuit is:
$I=\frac{V_G}{R_T+R_L}$
We wish to maximize power delivered to the load. Their ratio is simply:
$\frac{P_L}{P_T}=\frac{I^2R_L}{I^2R_T}=\frac{R_L}{R_T}$
And the efficiency:
$\eta=\frac{P_L}{P_T+P_L}=\frac{R_L}{R_T+R_L}$
So to make most power be used in the load, one must simply maximize $R_L$. But doing so would lower the delivered power:
$P_L=\frac{V_G^2R_L}{(R_T+R_L)^2}$
So we need to increase both $V_G$ and $R_L$ to maintain a constant power at the load $P_L$ while minimizing losses. This change in generator voltage and load impedance may both be accomplished by step-up and step-down transformers.
Note: The actual expression for the loss $P_T$ as a function of $V_G$, $P_L$ and $R_T$ is a bit complicated so I thought it's best to leave it out.
