Why does increasing resistance decrease the heat produced in an electric circuit? If $H=\frac{V^2}{R}{t}$  ,then increasing resistance means decreasing the heat produced. 
But, isnt it that the heat in a circuit is produced due to the presence of resistors? Moreover metals with high resistances are used as heating elements ,like Nichrome? 
Why does the equation state that the heat produced is inversely proportional to Resistance 
 A: Your statement 

If $\displaystyle{H=\frac{V^2}{R} t}$ ,then increasing resistance means decreasing the heat
  produced.

Implies that the voltage $V$ stays constant.
So with $V=IR$ if $V$ stays constant and the resistance $R$ increases then the current $I$ decreases.  
A classic example of this happening is in a tungsten filament light bulb  
When the bulb is first switched on a current flows through the filament and the power dissipated is $\displaystyle{\frac {V^2}{R}}$.
As the filament heats up the current flowing through the filament decreases because of the increased resistance of the filament and so the power dissipated decreases.
This larger current flowing through the filament is a reason why filament light bulbs often blow just as they are switched on.
Update in answer to a comment
Remembering that $V=IR$ then for a constant voltage if the resistance $R$ goes up by a factor $k$ then the current goes down by a factor $k$.
Power = $I^2 R$ so if the resistance $R$ has increased by a factor $k$ to $kR$ and the current has decreased to $\dfrac I k$ then the power is now $\left ( \dfrac {I} {k} \right )^2 kR = \dfrac {I^2R}{k}$.
This means that the electrical power dissipated has decreased by a facor $k$ when the resistance has increased by a factor of $k$.
A: Yes, there are two separate issues that involve resistance to keep track of.  The first is, what is the current that will run through the circuit, given the voltage.  That depends on the resistance such that the lower the resistance, the higher the current, and that's where the counterintuitive behavior is coming into play when you look at the heat generated.  But the second question is, where is that heat generated given that you already know the current, and this is the perfectly intuitive part-- the heat comes from the highest contribution to the resistance.  So the reason you get lots of heat from a smaller resister is only because the resistance in the rest of the circuit is very low-- it's no longer true if you put something in that has even less resistance than what the rest of the circuit already has.
A: You are just messing up with the equations. In order to state a proportionality relationship between a physical quantity and some other quantities, you need to be make sure that the quantities are all independent.   
In your equation, the voltage and resistance are not independent quantities, but voltage is a function of resistance and the current flowing through it. Hence to state a relationship between resistance and heat generated, you need to crack down the voltage into independent quantities- resistance and current. Then the equation reads $\displaystyle{H=I^2Rt}$, which tells us that heat is directly proportional to resistance.   
The source of your confusion was that you used voltage, which depends on current and resistance (defining relation between heat and resistance with a quantity involving resistance) to define the proportionality relation between heat and resistance.
A: Consider analogy with fluid flow. When there flow in a pipe say, due to friction, energy of motion is dissipated away into heat. Therefore for dissipation into heat to occur two things are necessary: flow, and resistance to flow. In the absence of either of them there is no dissipation into heat. Same is true of current in a circuit.
In the equation $H=\frac{V^2}{R}t$ you can only see what the resistance is, but not what the current is. Turns out that for constant applied voltage, if you increase resistance, current reduces more than proportionately (see @Farcher's answer). In the extreme case if there is break in the circuit and therefore no current, there would be no heat generation at all.
A: Agreed it is confusing, but consider two ohmic resistors, where resistor 1 is half of the value of resistor 2.  And let us say that the same voltage is applied across both resistors.  Now since the current through each resistor is inversely proportional to the resistance, then current in resistor 1 is twice the current in resistor 2, and the heat dissipated in resistor 1 is twice the heat dissipated in resistor 2 since heat dissipated, $H=IVt$.  Same conclusion is reached by using $H=I^2Rt$
