Finding value of an algebraic expression. I am given $$2x = a-\frac{1}{a} \;\text{and} \; 2y=b-\frac{1}{b} .$$
I have to find the value of $ xy +\sqrt{(x²+1)(y²+1)}.$
One way to solve this is to simply putting the value of $x$ and $y$ and get the answer, but that's a very lengthy process though lucid. But I am looking for a short cut method or some kind of tricks to solve the problem.
If anyone can visualize it, I wil be grateful for sharing.
Thank you.
 A: I do not think there is any special trick in solving the problem. 
$$x=\frac{a}{2}-\frac{1}{2a}$$
$$y=\frac{b}{2}-\frac{1}{2b}$$
$$xy=\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab}$$
$$x^2+1 = \frac{a^2}{4}-\frac{1}{2}+\frac{1}{4a^2}+1=\frac{a^2}{4}+\frac{1}{2}+\frac{1}{4a^2}=(\frac{a}{2}+\frac{1}{2a})^2$$
$$y^2+1 = \frac{b^2}{4}-\frac{1}{2}+\frac{1}{4b^2}+1=\frac{b^2}{4}+\frac{1}{2}+\frac{1}{4b^2}=(\frac{b}{2}+\frac{1}{2b})^2$$
Putting in the above values
$$xy+\sqrt{(x^2+1)(y^2+1)}=(\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab})+\sqrt{(\frac{a}{2}+\frac{1}{2a})^2*(\frac{b}{2}+\frac{1}{2b})^2}$$
Assuming that $=\frac{a}{2}+\frac{1}{2a}>0$ and $\frac{b}{2}+\frac{1}{2b}>0$
$$=(\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab})+(\frac{a}{2}+\frac{1}{2a})*(\frac{b}{2}+\frac{1}{2b})$$
$$=(\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab})+(\frac{ab}{4}+\frac{a}{4b}+\frac{b}{4a}+\frac{1}{4ab})$$
$$=\frac{ab}{4}+\frac{1}{4ab}+\frac{ab}{4}+\frac{1}{4ab}$$
So, 
$$xy+\sqrt{(x^2+1)(y^2+1)}=(\frac{ab}{2}+\frac{1}{2ab})$$
A: Take $t_1=a-\frac1a$ and $t_2=b-\frac1b$ and then solve
$$x=\frac{t_1}{2},y=\frac{t_2}{2}$$
Now $$xy+\sqrt{(x^2+1)(y^2+1)}$$
$$\left(\frac{t_1}{2}\right)\left(\frac{t_2}{2}\right)+\sqrt{\left(\left(\frac{t_1}{2}\right)^2+1\right)\left(\left(\frac{t_2}{2}\right)^2+1\right)}$$
$$=\frac{t_1t_2}{4}+\sqrt{\frac{t_1^2+4}{4}\cdot \frac{t_2^2+4}{4}}$$
$$=\frac{t_1t_2}{4}+\frac{\sqrt{(t_1^2+4)(t_2^2+4)}}{4}$$
and then substitute back $t_1=a-\frac1a,t_2=b-\frac1b$
