A) \[\frac{1}{13}\]
B) 3
C) \[\frac{1}{2}\]
D) 2
Correct Answer: B
Solution :
| \[\because \,\,{{\cot }^{-\,1}}(7)+{{\cot }^{-\,1}}(8)+{{\cot }^{-\,1}}(18)=\theta \] |
| \[\Rightarrow {{\tan }^{-1}}\left( \frac{1}{7} \right)+{{\tan }^{-1}}\left( \frac{1}{8} \right)+{{\tan }^{-1}}\left( \frac{1}{18} \right)=\theta \]\[\Rightarrow {{\tan }^{-1}}\left( \frac{\frac{1}{7}+\frac{1}{8}}{1+\frac{1}{7}.\frac{1}{8}} \right)+{{\tan }^{-1}}\left( \frac{1}{18} \right)=\theta \]\[\Rightarrow {{\tan }^{-1}}\left( \frac{3}{11} \right)+{{\tan }^{-1}}\left( \frac{1}{18} \right)=\theta \] |
| \[\Rightarrow {{\tan }^{-1}}\left( \frac{65}{195} \right)=\theta \]\[\Rightarrow {{\tan }^{-1}}\left( \frac{1}{3} \right)=\theta \]\[\Rightarrow {{\cot }^{-1}}(3)=\theta \]\[\Rightarrow \cot \theta =3\] |
You need to login to perform this action.
You will be redirected in
3 sec
