Prove $\tan\big(\frac{\pi}{2} -\theta \big) = \cot \theta$ $$\tan\left(\frac{\pi}{2} -\theta \right) =  \cot \theta$$

I can prove this by changing into $\cos$ and $\sin$

But I want to know if it is possible to prove it using relation given below. If not why ?
$$\tan\left(\frac{\pi}{2} -\theta \right) =\frac{ \tan\frac{\pi}{2} -\tan\theta}{1+ \tan\frac{\pi}{2}\tan\theta} $$
No, it isn't possible with the identity in your question because $\tan \frac\pi2$ is not defined.  It isn't defined because $\cos \frac\pi 2 = 0$, and $\tan x = \frac{\sin x}{\cos x}$ in general.

Actually, it's exactly this fact ($\tan x = \frac{\sin x}{\cos x}$) that you'll want to use.  I'll get you started:

$$
  \tan\left(\frac{\pi}{2} - x\right) = \frac{\sin\left(\frac{\pi}{2} - x\right)}{\cos \left(\frac{\pi}{2} - x \right)}
   = \frac{\sin\frac\pi 2 \cos x - \cos \frac\pi 2 \sin x}{\cos \frac\pi 2 \cos x + \sin\frac \pi 2 \sin x}
$$

Of course it'd be simpler to use the cofunction identities $\sin(\frac \pi 2 - x) = \cos x$ and $\cos(\frac \pi 2 - x) = \sin x$, but I'm assuming that's not allowed (or you want something different) because your original question is basically asking for an explanation of the tangent cofunction identity.

Edit: Many other answers are mentioning limits, etc.  Technically those will work but I'm assuming you're currently a trig student, which means it's unlikely you've had formal exposure to limits, and especially to l'Hôpital's rule.
If instead of $$\ tan (\frac{\pi}{2} - x)$$ We take 
$$ \lim_ {a \to \pi/2} \ tan(a-x)$$
Then $$\tan(a-x) = \frac{(\tan a  - \tan x)}{ (1+\tan (a)\tan(x))}$$ 
Divide by $\tan(a)$ and use $\frac{1}{\tan(a)}=0$