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.

Simplify the numerator. You will see immediately how the result was obtained. — Math Lover, Nov 29 '17 at 15:36

score 3 · Accepted Answer · edited Nov 29 '17 at 15:42

3

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

edited Nov 29 '17 at 15:42

Math Lover

15,153

answered Nov 29 '17 at 15:37

Botond

11,938

Thanks @MathLover, I thought it would not fit well with the double dollar. – Botond Nov 29 '17 at 15:44

Fraction transformation

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