What is the inverse of f(x)=tan(x/(1+x)) ?

The inverse of f(x)=tan(x/(1+x)) is (arctan(x))/(1-arctan(x))

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inverse of f(x)=tan(x/(1+x))

inverse

f (x) = tan (\frac{x}{1 + x})

\frac{arctan ( x )}{1 - arctan ( x )}

Solution steps

tan (a) = b \Rightarrow a = arctan (b)

\frac{x}{1 + x} = arctan (y)

Solve for

x, \frac{x}{1 + x} = arctan (y)

\frac{arctan ( y )}{1 - arctan ( y )}

Substitute

y = x

\frac{arctan ( x )}{1 - arctan ( x )}

f (x) = \frac{9 + 10 x}{8}

f (x) = \frac{6 x}{x - 3}, x \neq = 3

y = 43 x + 2

f (x) = 1.35 x + 114 - 2.85 x

What is the inverse of f(x)=tan(x/(1+x)) ?
The inverse of f(x)=tan(x/(1+x)) is (arctan(x))/(1-arctan(x))