# Examples on Inverse Trigonometry Set 2

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Example-79

If  $${\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z = \pi ,$$  find the value of  $${x^2} + {y^2} + {z^2} + 2xyz$$

Solution:

\begin{align} {\cos ^{ - 1}}x + {\cos ^{ - 1}}y&= \pi - {\cos ^{ - 1}}z \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&= {\cos ^{ - 1}}( - z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{How?}}} \right)\\ \end{align}

\begin{align}&\Rightarrow \quad {\cos ^{ - 1}}\left( {xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right) = {\cos ^{ - 1}}( - z) \\ &\Rightarrow \quad xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} = - z \\ &\Rightarrow\quad {(xy + z)^2} = (1 - {x^2})(1 - {y^2}) \\ \end{align}

Rearranging this yields the required value:

${x^2} + {y^2} + {z^2} + 2xyz = 1$

Example-80

Evaluate the sum of the following series:

$S = {\tan ^{ - 1}}\frac{1}{{2 \cdot {1^2}}} + {\tan ^{ - 1}}\frac{1}{{2 \cdot {2^2}}} + {\tan ^{ - 1}}\frac{1}{{2 \cdot {3^2}}} + ....\infty$

Solution: The approach in such questions is what we mentioned earlier: write each term in the series as a difference, so that successive terms cancel out..

The general $${r^{{\text{th}}}}$$ term in S is

${T_r} = {\tan ^{ - 1}}\frac{1}{{2 \cdot {r^2}}}$

We know that \begin{align}{\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}x - {\tan ^{ - 1}}y\end{align}. Somehow, we have to use this term to express $${T_r}$$ as a difference. To do that we express the denominator as $$1 + xy:$$

\begin{align} \frac{1}{{2 \cdot {r^2}}} = \frac{2}{{4 \cdot {r^2}}} = \frac{2}{{1 + (4{r^2} - 1)}}&= \frac{2}{{1 + (2r + 1)(2r - 1)}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{{(2r + 1) - (2r - 1)}}{{1 + (2r + 1)(2r - 1)}}\\ \end{align}

This means that

${T_r} = {\tan ^{ - 1}}(2r + 1) - {\tan ^{ - 1}}(2r - 1)$

Expressing $${T_r}$$ this way solves our problem, since S now becomes

$S = ({\tan ^{ - 1}}3 - {\tan ^{ - 1}}1) + ({\tan ^{ - 1}}5 - {\tan ^{ - 1}}3) + ({\tan ^{ - 1}}7 - {\tan ^{ - 1}}5) + ...({\tan ^{ - 1}}(2n + 1) - {\tan ^{ - 1}}(2n - 1))$

$${\text{where}}\;n \to \infty$$. Thus,

\begin{align}&S = \mathop {\lim }\limits_{n \to \infty } \;{\tan ^{ - 1}}(2n + 1) - {\tan ^{ - 1}}1 \\ \,\,\,\, &\quad= \frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4} \\ \end{align}

Example-81

Find the sum  \begin{align}S = \sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}} \left( {\frac{{2r}}{{{r^4} + {r^2} + 2}}} \right)\end{align}

Solution: The $${r^{th}}$$ term is

\begin{align}&{T_r} = {\tan ^{ - 1}}\left( {\frac{{2r}}{{{r^4} + {r^2} + 2}}} \right) \\ &\;\;\;= {\tan ^{ - 1}}\left( {\frac{{({r^2} + r + 1) - ({r^2} - r + 1)}}{{1 + ({r^2} + r + 1)({r^2} - r + 1)}}} \right) \\ &\;\;\;= {\tan ^{ - 1}}({r^2} + r + 1) - {\tan ^{ - 1}}({r^2} - r + 1) \\ \end{align}

Therefore,

$S = \left( {{{\tan }^{ - 1}}3 - {{\tan }^{ - 1}}1} \right) + \left( {{{\tan }^{ - 1}}7 - {{\tan }^{ - 1}}3} \right) + ... + \left( {{{\tan }^{ - 1}}({n^2} + n + 1) - {{\tan }^{ - 1}}({n^2} - n + 1)} \right)$

where  $$n \to \infty$$

\begin{align}&\Rightarrow \quad S = \mathop {\lim }\limits_{n \to \infty } \;\,{\tan ^{ - 1}}({n^2} + n + 1) - {\tan ^{ - 1}}1 \\ &\quad\qquad\;= \frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4}\\ \end{align}

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