Question

If f(y) = $\frac{y}{\sqrt{1 - y^{2}}}$, g(y) = $\frac{y}{\sqrt{1 + y^{2}}}$ then show that (fog) (y) = y.

Answer 1

Answer:

(fog)(y) = y

Step-by-step explanation:

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Answer 2

Solution:

$f(y) = \frac{y}{ \sqrt{1 - {y}^{2} } } \\ \\ g(y) = \frac{y}{ \sqrt{1 + {y}^{2} } }$
for fog(y) put the value of g(y) into f(y)

I.e.
$f(g(y)) = \frac{ \frac{y}{ \sqrt{1 + {y}^{2} } } }{ \sqrt{1 - { (\frac{y}{ \sqrt{1 + {y}^{2} } } )}^{2} } } \\ \\ = \frac{ \frac{y}{ \sqrt{1 + {y}^{2} } } }{ \sqrt{1 - { (\frac{{y}^{2}}{1 + {y}^{2} } )}} } \\ \\ \\ = \frac{ \frac{y}{ \sqrt{1 + {y}^{2} } } }{ \sqrt{ { \frac{{ 1 - {y}^{2} + y}^{2}}{1 + {y}^{2} } }} } \\ \\ = \frac{ \frac{y}{ \sqrt{1 + {y}^{2} } } }{ \frac{1}{ \sqrt{1 + {y}^{2} } } } \\ \\ = y \\ \\ fog(y) = y$