# 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}