A) 0
B) \[\frac{\pi }{4}\]
C) \[{{\tan }^{-1}}2\]
D) \[\frac{\pi }{2}\]
Correct Answer: B
Solution :
Key Idea: \[\cos e{{c}^{-1}}a={{\sin }^{-1}}\frac{1}{a}\] and \[{{\sin }^{-1}}a={{\cot }^{-1}}\left( \frac{\sqrt{(1-{{a}^{2}})}}{a} \right)\] Given \[{{\cot }^{-1}}9+\cos e{{c}^{-1}}\frac{\sqrt{41}}{4}\] \[={{\cot }^{-1}}9+{{\sin }^{-1}}\left( \frac{4}{\sqrt{41}} \right)\] \[={{\cot }^{-1}}9+{{\cot }^{-1}}\left[ \frac{\sqrt{1-\frac{16}{41}}}{4/\sqrt{41}} \right]\] \[={{\cot }^{-1}}9+{{\cot }^{-1}}\left[ \frac{5}{4} \right]\] \[={{\tan }^{-1}}\left( \frac{1}{9} \right)+{{\tan }^{-1}}\left( \frac{4}{5} \right)\] \[={{\tan }^{-1}}\left( \frac{\frac{1}{9}+\frac{4}{5}}{1-\frac{1}{9}.\frac{4}{5}} \right)={{\tan }^{-1}}\left( \frac{5+36}{45-4} \right)\] \[={{\tan }^{-1}}\left( \frac{41}{41} \right)\] \[={{\tan }^{-1}}(1)=\frac{\pi }{4}\] \[\therefore \] \[{{\cot }^{-1}}9+\cos e{{c}^{-1}}=\frac{\sqrt{41}}{4}=\frac{\pi }{4}\]
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