If  f(x)=x+tanx and f is inverse of g, then  g1(x) is equal to

# If  $f\left(x\right)=x+\mathrm{tan}x$ and f is inverse of g, then  ${g}^{1}\left(x\right)$ is equal to

1. A

$\frac{1}{1+{\left[g\left(x\right)-x\right]}^{2}}$

2. B

$\frac{1}{1-{\left[g\left(x\right)-x\right]}^{2}}$

3. C

$\frac{1}{2+{\left[g\left(x\right)-x\right]}^{2}}$

4. D

$\frac{1}{2-{\left[g\left(x\right)-x\right]}^{2}}$

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### Solution:

$g\left(x\right)={f}^{-1}\left(x\right)⇒f\left(g\left(x\right)\right)=x$

$⇒{f}^{|}\left(g\left(x\right)\right){g}^{|}\left(x\right)=1$

$⇒{g}^{|}\left(x\right)=\frac{1}{{f}^{|}\left(g\left(x\right)\right)}$

$=\frac{1}{1+{\mathrm{sec}}^{2}\left(g\left(x\right)\right)}$

$=\frac{1}{1+{\mathrm{tan}}^{2}g\left(x\right)+1}$

$=\frac{1}{2+\left[f\left(g\left(x\right)\right)\right]-{\left(g\left(x\right)\right)}^{2}}$

$=\frac{1}{2+{\left(x-g\left(x\right)\right)}^{2}}$

$=\frac{1}{2+{\left(g\left(x\right)-x\right)}^{2}}$  +91

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