Interconversion Between Inverse Trigonometric Ratios

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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

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

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