A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[{{\cot }^{-1}}5\]
D) \[{{\cot }^{-1}}3\]
Correct Answer: D
Solution :
[d] We have \[{{\cot }^{-1}}7+{{\cot }^{-1}}8+{{\cot }^{-1}}18\] \[{{\tan }^{-1}}\frac{1}{7}+{{\tan }^{-1}}\frac{1}{8}+{{\tan }^{-1}}\frac{1}{18}\] \[={{\tan }^{-1}}\left( \frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\times \frac{1}{8}} \right)+{{\tan }^{-1}}\frac{1}{18}\] \[={{\tan }^{-1}}\frac{15}{55}{{\tan }^{-1}}\frac{1}{18}\left( \because \frac{1}{7}.\frac{1}{8}<1 \right)\] \[{{\tan }^{-1}}\frac{3}{11}+{{\tan }^{-1}}\frac{1}{18}={{\tan }^{-1}}\left( \frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11}\times \frac{1}{18}} \right)\] \[\left( \because \frac{3}{11}.\frac{1}{18}<1 \right)\] \[={{\tan }^{-1}}\frac{65}{195}={{\tan }^{-1}}\frac{1}{3}={{\cot }^{-1}}3\]
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