What is the inverse of y=arctan(1/(8x+9)) ?

The inverse of y=arctan(1/(8x+9)) is (1-9tan(x))/(8tan(x))

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inverse of y=arctan(1/(8x+9))

inverse

y = arctan (\frac{1}{8 x + 9})

\frac{1 - 9 tan ( x )}{8 tan ( x )}

Solution steps

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

\frac{1}{8 x + 9} = tan (y)

Solve for

x, \frac{1}{8 x + 9} = tan (y)

\frac{1 - 9 tan ( y )}{8 tan ( y )}

Substitute

y = x

\frac{1 - 9 tan ( x )}{8 tan ( x )}

f (x) = e^{2 \cdot (5 x + 3)}

f (x) = 4 - x^{2}, - 2 \leq x < 0

\frac{1}{x + 7}

y = \frac{5}{2} x - 4

f (x) = x - 5 + 9

What is the inverse of y=arctan(1/(8x+9)) ?
The inverse of y=arctan(1/(8x+9)) is (1-9tan(x))/(8tan(x))