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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