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