# Relation f⁻¹(x) with f⁻¹(1/x) – 2

## Relation f⁻¹(x) with f⁻¹(1/x) – 2

Theorem: cos⁻¹(1/x) = sec⁻¹(x), for all $$x\in\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)$$

Proof: cos⁻¹(1/x) = sec⁻¹(x)

Let sec⁻¹(x) = θ

Where θ ϵ [0, π] –{0} and $$x\in\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)$$

x = secθ

1/x = 1/ secθ

1/x = cosθ

θ = cos⁻¹(1/x)

sec⁻¹(x) = θ

sec⁻¹(x) = cos⁻¹(1/x)

cos⁻¹(1/x) = sec⁻¹(x), for all $$x\in\left( -\infty ,-1 \right]\cup \left( 1,\infty \right]$$

Hence proved

Theorem: {{\tan }^{-1}}\left( \frac{1}{x} \right)=\left\{ \begin{align}& {{\cot }^{-1}}x,\ \ \ \ \ \ \ \ \ \ \ for\ \ x>0 \\ & -\pi +{{\cot }^{-1}}x,\ \ \ \ for\ \ x<0 \\ \end{align} \right.

Proof: tan⁻¹(1/x) = cot⁻¹(x)

Let cot⁻¹(x) = θ

1/x = 1/ cotθ

1/x = tanθ

θ = tan⁻¹(1/x)

cot⁻¹(x) = θ

cot⁻¹(x) = tan⁻¹(1/x)

{{\tan }^{-1}}\left( \frac{1}{x} \right)=\left\{ \begin{align} & \theta ,\ \ \ \ \ \ \ \ \ \ \ for\ \ 0<\theta <\frac{\pi }{2} \\ & -\pi +\theta ,\ \ \ \ for\ \ \frac{\pi }{2}<\theta <\pi \\ \end{align} \right.,

=\left\{ \begin{align} & co{{t}^{-1}}\left( x \right),\ \ \ \ \ \ \ \ \ \ \ for\ \ 0<co{{t}^{-1}}\left( x \right)<\frac{\pi }{2} \\ & -\pi +co{{t}^{-1}}\left( x \right)\ \ \ \ for\ \ \frac{\pi }{2}<co{{t}^{-1}}\left( x \right)<\pi \\ \end{align} \right.,

{{\tan }^{-1}}\left( \frac{1}{x} \right)=\left\{ \begin{align} & {{\cot }^{-1}}x,\ \ \ \ \ \ \ \ \ \ \ for\ \ x>0 \\ & -\pi +{{\cot }^{-1}}x,\ \ \ \ for\ \ x<0 \\ \end{align} \right.,

Hence proved