A) \[\frac{1}{13}\]
B) 3
C) \[\frac{1}{2}\]
D) 2
Correct Answer: B
Solution :
\[\because \] \[{{\cot }^{-1}}\,(7)+co{{t}^{-1}}\,(8)+co{{t}^{-1}}\,(18)=\theta \] |
\[\because \] \[\Rightarrow \]\[{{\tan }^{-1}}\left( \frac{1}{7} \right)+{{\tan }^{-1}}\left( \frac{1}{8} \right)+{{\tan }^{-1}}\left( \frac{1}{18} \right)=\theta \] |
\[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\cdot \frac{1}{8}} \right)+{{\tan }^{-1}}\left( \frac{1}{18} \right)=\theta \] |
\[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{3}{11} \right)+{{\tan }^{-1}}\left( \frac{1}{18} \right)=\theta \] |
\[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{65}{195} \right)=\theta \] \[\Rightarrow \]\[{{\tan }^{-1}}\left( \frac{1}{3} \right)=\theta \] |
\[\Rightarrow \]\[{{\cot }^{-1}}\,(3)=\theta \] |
\[\Rightarrow \]\[{{\cot }^{-1}}\theta =3\] |
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