A) \[0\]
B) \[{{\tan }^{-1}}2\]
C) \[\pi /4\]
D) \[\pi /2\]
Correct Answer: C
Solution :
\[{{\cot }^{-1}}9+\text{cose}{{\text{c}}^{-1}}\frac{\sqrt{41}}{4}\] \[={{\cot }^{-1}}9+{{\sin }^{-1}}\frac{4}{\sqrt{41}}={{\cot }^{-1}}9+{{\tan }^{-1}}\frac{4}{5}\] \[={{\tan }^{-1}}(1/9)+{{\tan }^{-1}}(4/5)\] \[={{\tan }^{-1}}\left| \frac{1/9+4/5}{1-1/9\times 4/5} \right|={{\tan }^{-1}}\left( \frac{5+36}{45-4} \right)={{\tan }^{-1}}\left( \frac{41}{41} \right)\]\[={{\tan }^{-1}}(1)=\pi /4.\]
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