If xy + yz + zx = 1, ........... If $xy + yz + zx = 1$, then show that 

$$\dfrac{x}{1-x^2} + \dfrac{y}{1-y^2} + \dfrac{z}{1-z^2} = \dfrac{4xyz}{(1-x^2)(1-y^2)(1-z^2)}$$

I tried doing the sum algebraically, that is, by solving the LHS but, that method isn't really getting us anywhere.
Also, I found this sum in an a book on trigonometry. So how can I solve it with trigonometry?

For a trigonometric proof, follow Parth Kohli's suggestion which is nice. Meanwhile here's an algebraic proof.
Let $a=\dfrac{x}{1-x^2}$, $b=\dfrac{y}{1-y^2}$, $c=\dfrac{z}{1-z^2}$.
We need to prove: $a+b+c=4abc$, i.e. $\sum \dfrac{1}{ab}=4$.
We have $\sum \dfrac{1}{ab}=\sum \dfrac{(1-x^2)(1-y^2)}{xy}=1 + \sum \left( \dfrac{1}{xy}-(\dfrac{x}{y}+\dfrac{y}{x}) \right)$.
Now, $\sum \dfrac{1}{xy}=\sum \dfrac{xy+yz+zx}{xy}=3+\sum (\dfrac{x}{y}+\dfrac{y}{x})$, 
which gives the desired equality.
