Fraction transformation

Question

I am not sure how it is possible to write $\frac{\frac{t}{\sqrt{t^2+1}}+1}{\sqrt{t^2+1}+t}$ as $\frac{1}{\sqrt{t^2+1}}$ Does anyone know how I can transform this? I would appriciate a step-by-step solution.

Answer 1

$$\frac{\frac{t}{\sqrt{t^2+1}}+1}{\sqrt{t^2+1}+t}=\frac{\frac{t}{\sqrt{t^2+1}}+\frac{\sqrt{t^2+1}}{\sqrt{t^2+1}}}{\sqrt{t^2+1}+t}=\frac{\frac{t+\sqrt{t^2+1}}{\sqrt{t^2+1}}}{\sqrt{t^2+1}+t}=\frac{t+\sqrt{t^2+1}}{\sqrt{t^2+1}}\cdot\frac{1}{\sqrt{t^2+1}+t}=\frac{1}{\sqrt{t^2+1}}$$