# Interconversion Between Inverse Trigonometric Ratios

The following sets of conversions should be fairly obvious:

(a)  \begin{align}{\sin ^{ - 1}}x = {\cos ^{ - 1}}\sqrt {1 - {x^2}} = {\tan ^{ - 1}}\left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right) = {\cot ^{ - 1}}\left( {\frac{{\sqrt {1 - {x^2}} }}{x}} \right) = {\sec ^{ - 1}}\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right) = {\text{cose}}{{\text{c}}^{ - 1}}\left( {\frac{1}{x}} \right)\end{align}

(b) \begin{align}{\cos ^{ - 1}}x = {\sin ^{ - 1}}\sqrt {1 - {x^2}} = {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 - {x^2}} }}{x}} \right) = {\cot ^{ - 1}}\left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right) = {\sec ^{ - 1}}\frac{1}{x} = {\text{cose}}{{\text{c}}^{ - 1}}\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)\end{align}

(c)  \begin{align}{\tan ^{ - 1}}x = {\sin ^{ - 1}}\left( {\frac{x}{{\sqrt {1 + {x^2}} }}} \right) = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {x^2}} }}} \right) = {\cot ^{ - 1}}\left( {\frac{1}{x}} \right) = {\sec ^{ - 1}}\sqrt {1 + {x^2}} = {\text{cose}}{{\text{c}}^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} }}{x}} \right)\end{align}  etc.

This properties follow from definitions. For example:

We also have relations relating  $$2{\tan ^{ - 1}}x$$  to  $${\sin ^{ - 1}}$$  and $${\cos ^{ - 1}}$$ :

Let  $${\tan ^{ - 1}}x = \theta \quad \Rightarrow x = \tan \theta$$      such that  \begin{align}\theta \in \left( {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right)\end{align}

Now,

$\sin 2\theta = \frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \frac{{2x}}{{1 + {x^2}}}$

 If  \begin{align}2\theta \in \left( {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right)\end{align} : \begin{align}\Rightarrow \quad \theta \in \left( {\frac{{ - \pi }}{4},\;\frac{\pi }{4}} \right) \;\;\;\; &\Rightarrow \quad \;\;\;x = \tan \theta \in ( - 1,\;1)\, \\ &\Rightarrow\quad 2\theta = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) \\ \end{align} If   \begin{align}2\theta > \frac{\pi }{2}\end{align} : \begin{align}\Rightarrow \quad x > 1 \;\;\; \Rightarrow \quad \;\;2\theta = \pi - {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\end{align} If   \begin{align}2\theta < \frac{{ - \pi }}{2}\end{align} : \begin{align}\Rightarrow \; x < - 1 \;\;\;\;\;\quad \Rightarrow \quad\;\;\;\;\;2\theta = - \pi - {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\end{align}

Thus,

2{\tan ^{ - 1}}x = \left\{ \begin{align} {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right), \quad x \in [ - 1,\;1] \\\\pi - {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right), x > 1 \\ - \pi - {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right), x < - 1 \\ \end{align} \right\}

Following a similar approach, show that

2{\tan ^{ - 1}}x = \left\{ \begin{align} {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right), x \in \left[ {0,\;\infty } \right) \\ - {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right), x \in \left( { - \infty ,\;0} \right] \\ \end{align} \right\}