If $$\frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}+\frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}}$$= 62, then what is the value of $$x$$ ($$x < 0$$) ?

Expression : $$\frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}+\frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}}=62$$

=> $$\frac{(x+\sqrt{x^2-1})^2+(x-\sqrt{x^2-1})^2}{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}=62$$

=> $$\frac{(x^2+x^2-1+2x\sqrt{x^2-1})+(x^2+x^2-1-2x\sqrt{x^2-1})}{(x^2)-(x^2-1)}=62$$

=> $$\frac{4x^2-2}{1}=62$$

=> $$4x^2-2=62$$

=> $$4x^2=62+2=64$$

=> $$x^2=\frac{64}{4}=16$$

=> $$x=\sqrt{16}=\pm4$$

$$\because x<0$$, => $$x=-4$$

=> Ans - (A)