Examples on Inverse Trigonometry Set 1

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Example-76

Find the value of

(a) \(\begin{align}{S_1} = {\tan ^{ - 1}}\frac{1}{2} + {\tan ^{ - 1}}\frac{1}{5} + {\tan ^{ - 1}}\frac{1}{8}\end{align}\)             (b)  \(\begin{align}{S_2} = {\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{5} + {\tan ^{ - 1}}\frac{1}{7} + {\tan ^{ - 1}}\frac{1}{8}\end{align}\)

Solution: (a)

\[\begin{align}&{S_1} = \left( {{{\tan }^{ - 1}}\frac{1}{2} + {{\tan }^{ - 1}}\frac{1}{5}} \right) + {\tan ^{ - 1}}\frac{1}{8} \\  &\quad= {\tan ^{ - 1}}\left( {\frac{{\frac{1}{2} + \frac{1}{5}}}{{1 - \frac{1}{2} \times \frac{1}{5}}}} \right) + {\tan ^{ - 1}}\frac{1}{8} \\ &\quad= {\tan ^{ - 1}}\frac{7}{9} + {\tan ^{ - 1}}\frac{1}{8}  \\   &\quad= {\tan ^{ - 1}}\left\{ {\frac{{\frac{7}{9} + \frac{1}{8}}}{{1 - \frac{7}{9} \times \frac{1}{8}}}} \right\} = {\tan ^{ - 1}}\left( {\frac{{65}}{{65}}} \right) = {\tan ^{ - 1}}1 = \frac{\pi }{4}  \\ \end{align} \]

(b)

\[\begin{align}&{S_2} = \left( {{{\tan }^{ - 1}}\frac{1}{3} + {{\tan }^{ - 1}}\frac{1}{7}} \right) + {\tan ^{ - 1}}\frac{1}{5} + {\tan ^{ - 1}}\frac{1}{8}  \\   &\quad= {\tan ^{ - 1}}\frac{1}{2} + {\tan ^{ - 1}}\frac{1}{5} + {\tan ^{ - 1}}\frac{1}{8}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{verify!}}} \right) \\   &\quad= {S_1} = \frac{\pi }{4} \\ \end{align} \]

Example-77

Evaluate  \(\begin{align}S = {\sin ^{ - 1}}\frac{{12}}{{13}} + {\cos ^{ - 1}}\frac{4}{5} + {\tan ^{ - 1}}\frac{{63}}{{16}}\end{align}\)

Solution: We convert both  \({\sin ^{ - 1}}\) and \({\cos ^{ - 1}}\) to \({\tan ^{ - 1}}\) :

\[S = {\tan ^{ - 1}}\frac{{12}}{5} + {\tan ^{ - 1}}\frac{3}{4} + {\tan ^{ - 1}}\frac{{63}}{{16}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{How?}}} \right)\]

Note that  \(\begin{align}\frac{{12}}{5} \times \frac{3}{4} > 1,\end{align}\)  so

\[\begin{align} {\tan ^{ - 1}}\frac{{12}}{5} + {\tan ^{ - 1}}\frac{3}{4} &= \pi  + {\tan ^{ - 1}}\left( {\frac{{\frac{{12}}{5} + \frac{3}{4}}}{{1 - \frac{{12}}{5} \times \frac{3}{4}}}} \right)  \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \pi  - {\tan ^{ - 1}}\frac{{63}}{{16}} \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow\quad &S = \pi \\ \end{align} \]

Example-78

Evaluate  \(\begin{align}S = 2{\tan ^{ - 1}}\frac{1}{5} + {\sec ^{ - 1}}\frac{{5\sqrt 2 }}{7} + 2{\tan ^{ - 1}}\frac{1}{8}\end{align}\)

Solution:

\[\begin{align}2\left\{ {{{\tan }^{ - 1}}\frac{1}{5} + {{\tan }^{ - 1}}\frac{1}{8}} \right\} &= 2{\tan ^{ - 1}}\left( {\frac{{\frac{1}{5} + \frac{1}{8}}}{{1 - \frac{1}{5} \times \frac{1}{8}}}} \right) \\   &= 2{\tan ^{ - 1}}\frac{1}{3}  \\   &= {\tan ^{ - 1}}\left( {\frac{{2 \times \frac{1}{3}}}{{1 - {{\left( {\frac{1}{3}} \right)}^2}}}} \right) = {\tan ^{ - 1}}\frac{3}{4} \\ \end{align} \]

Also,  \(\begin{align}{\sec ^{ - 1}}\frac{{5\sqrt 2 }}{7} = {\tan ^{ - 1}}\sqrt {{{\left( {\frac{{5\sqrt 2 }}{7}} \right)}^2} - 1}  = {\tan ^{ - 1}}\frac{1}{7}\end{align}\)

Thus,

\[\begin{align}S = {\tan ^{- 1}}\frac{3}{4} + {\tan ^{ - 1}}\frac{1}{7}&= {\tan ^{ - 1}}\left( {\frac{{\frac{3}{4} + \frac{1}{7}}}{{1 - \frac{3}{4} \times \frac{1}{7}}}} \right)  \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= {\tan ^{ - 1}}\;1  \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{\pi }{4}  \\ \end{align} \]

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