What is the inverse of 1/(sqrt(1+x^2)) ?

The inverse of 1/(sqrt(1+x^2)) is (sqrt(1-x^2))/x ,-(sqrt(1-x^2))/x

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inverse of 1/(sqrt(1+x^2))

inverse

\frac{1}{1 + x ^{2}}

\frac{1 - x ^{2}}{x}, - \frac{1 - x ^{2}}{x}

Solution steps

y = \frac{1}{1 + x ^{2}}

Replace

x

with

y

x = \frac{1}{1 + y ^{2}}

Solve for

y, x = \frac{1}{1 + y ^{2}}

\frac{1 - x ^{2}}{x}, - \frac{1 - x ^{2}}{x}

f (x) = \frac{8}{4 + x}

\frac{s + 1}{( s - 1 ) ^{2} + 1}

y = \frac{1}{6} \cdot 1 0^{x - 1} + 5

f (x) = - x + 1

f (x) = x^{5} + 5

What is the inverse of 1/(sqrt(1+x^2)) ?
The inverse of 1/(sqrt(1+x^2)) is (sqrt(1-x^2))/x ,-(sqrt(1-x^2))/x