A) \[{{\tan }^{-1}}\frac{y}{x}\]
B) \[{{\tan }^{-1}}yx\]
C) \[{{\tan }^{-1}}\frac{x}{y}\]
D) \[{{\tan }^{-1}}(x-y)\]
Correct Answer: C
Solution :
\[{{\tan }^{-1}}\left( \frac{{{c}_{1}}x-y}{{{c}_{1}}y+x} \right)+{{\tan }^{-1}}\left( \frac{{{c}_{2}}-{{c}_{1}}}{1+{{c}_{2}}{{c}_{1}}} \right)+{{\tan }^{-1}}\left( \frac{{{c}_{3}}-{{c}_{2}}}{1+{{c}_{3}}{{c}_{2}}} \right)+\]\[.....+{{\tan }^{-1}}\frac{1}{{{c}_{n}}}\] = \[{{\tan }^{-1}}\left( \frac{\frac{x}{y}-\frac{1}{{{c}_{1}}}}{1+\frac{x}{y}.\frac{1}{{{c}_{1}}}} \right)+{{\tan }^{-1}}\left( \frac{\frac{1}{{{c}_{1}}}-\frac{1}{{{c}_{2}}}}{1+\frac{1}{{{c}_{1}}{{c}_{2}}}} \right)\]\[+{{\tan }^{-1}}\left( \frac{\frac{1}{{{c}_{2}}}-\frac{1}{{{c}_{3}}}}{1+\frac{1}{{{c}_{2}}{{c}_{3}}}} \right)+.......+{{\tan }^{-1}}\frac{1}{{{c}_{n}}}\] \[={{\tan }^{-1}}\frac{x}{y}-{{\tan }^{-1}}\frac{1}{{{c}_{1}}}+{{\tan }^{-1}}\frac{1}{{{c}_{1}}}-{{\tan }^{-1}}\frac{1}{{{c}_{2}}}+{{\tan }^{-1}}\frac{1}{{{c}_{2}}}\]\[-{{\tan }^{-1}}\frac{1}{{{c}_{3}}}+...+{{\tan }^{-1}}\frac{1}{{{c}_{n-1}}}-{{\tan }^{-1}}\frac{1}{{{c}_{n}}}+{{\tan }^{-1}}\frac{1}{{{c}_{n}}}\] =\[{{\tan }^{-1}}\left( \frac{x}{y} \right)\].
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