Examples on Inverse Trigonometry Set 3

Example-82

Find the sum \(\begin{align}S = \sum\limits_{r = 1}^\infty  {{{\cot }^{ - 1}}} \left( {{2^{r + 1}} + \frac{1}{{{2^r}}}} \right)\end{align}\)

Solution:

\[\begin{align}  {T_r} = {\cot ^{ - 1}}\left( {{2^{r + 1}} + \frac{1}{{{2^r}}}} \right)&= {\tan ^{ - 1}}\left( {\frac{{{2^r}}}{{{2^{2r + 1}} + 1}}} \right)  \\   &= {\tan ^{ - 1}}\left( {\frac{{{2^{r + 1}} - {2^r}}}{{1 + {2^r} \cdot {2^{r + 1}}}}} \right)  \\   &= {\tan ^{ - 1}}{2^{r + 1}} - {\tan ^{ - 1}}{2^r} \\ \end{align} \]

\[\begin{align}&\Rightarrow  \quad S = \frac{\pi }{2} - {\tan ^{ - 1}}2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{How?}}} \right) \\ &\qquad\quad= {\cot ^{ - 1}}2 \\ \end{align} \]

Example-83

Prove that  \(\begin{align}S = \cos \left[ {{{\tan }^{ - 1}}\left\{ {\sin ({{\cot }^{ - 1}}x)} \right\}} \right] = \sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}}\end{align} \)

Solution:  \(\begin{align}{\cot ^{ - 1}}x = {\sin ^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}\end{align}\)

\[\begin{align}   \Rightarrow  \qquad &S = \cos \left[ {{{\tan }^{ - 1}}\left\{ {\frac{1}{{\sqrt {1 + {x^2}} }}} \right\}} \right].\\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\;\;= \cos \left[ {{{\cos }^{ - 1}}\frac{1}{{\sqrt {1 + {{\left( {\frac{1}{{1 + {x^2}}}} \right)}^2}} }}} \right] \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\;\;= \sqrt {\frac{{1 + {x^2}}}{{2 + {x^2}}}} \\ \end{align} \]

Example-84

Find the value of  \(\begin{align}S = {\cos ^{ - 1}}x + {\cos ^{ - 1}}\left\{ {\frac{x}{2} + \frac{{\sqrt {3 - 3{x^2}} }}{2}} \right\},\;{\text{when}}\;\frac{1}{2} \leqslant x \leqslant 1\end{align}\)

Solution: Since  \[x \in \left[ {\frac{1}{2},\;1} \right],\;{\cos ^{ - 1}}x \in \left[ {0,\;\frac{\pi }{3}} \right]\]. If  \({\cos ^{ - 1}}x = y,\) then

\[\begin{align}&S = y + {\cos ^{ - 1}}\left\{ {\frac{{\cos y}}{2} + \frac{{\sqrt 3 }}{2}\sin y} \right\}  \\  \,\,\,\, &\;\;= y + {\cos ^{ - 1}}\left\{ {\cos \left( {\frac{\pi }{3} - y} \right)} \right\}  \\ \end{align} \]

Since  \(\begin{align}y = {\cos ^{ - 1}}x \in \left[ {0,\;\frac{\pi }{3}} \right],\;\frac{\pi }{3} - y \in \left[ {0,\;\frac{\pi }{3}} \right]\end{align}\), so that

\[S = y + \frac{\pi }{3} - y = \frac{\pi }{3}\]

Example-85

Solve the equation  \(\begin{align}{\tan ^{ - 1}}\left( {\frac{{x - 1}}{{x - 2}}} \right) + {\tan ^{ - 1}}\left( {\frac{{x + 1}}{{x + 2}}} \right) = \frac{\pi }{4}\end{align}\)

Solution: If we try to combine the two  \({\tan ^{ - 1}}\) terms, we’ll get quadratic expressions. Instead, we write the  \(\begin{align}\frac{\pi }{4}\end{align}\)  on the RHS as  \({\tan ^{ - 1}}1\) and:

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

Therefore,

\[\frac{{x - 1}}{{x - 2}} = \frac{1}{{2x + 3}}\]

Solving this yields  \(\begin{align}x =  \pm \frac{1}{{\sqrt 2 }}\end{align}\)