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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.

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    $\begingroup$ Simplify the numerator. You will see immediately how the result was obtained. $\endgroup$ – Math Lover Nov 29 '17 at 15:36
  • $\begingroup$ ok, done thanks $\endgroup$ – Maddude Nov 29 '17 at 15:37
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$$\frac{\frac{t}{\sqrt{t^2+1}}+1}{\sqrt{t^2+1}+t}=\frac{\frac{t}{\sqrt{t^2+1}}+\frac{\sqrt{t^2+1}}{\sqrt{t^2+1}}}{\sqrt{t^2+1}+t}=\frac{\frac{t+\sqrt{t^2+1}}{\sqrt{t^2+1}}}{\sqrt{t^2+1}+t}=\frac{t+\sqrt{t^2+1}}{\sqrt{t^2+1}}\cdot\frac{1}{\sqrt{t^2+1}+t}=\frac{1}{\sqrt{t^2+1}}$$

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

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