A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) \[\pi \]
Correct Answer: B
Solution :
\[{{\cot }^{-1}}9+\text{cosec}{{\text{ }}^{-1}}\frac{\sqrt{41}}{4}\] \[={{\cot }^{-1}}9+{{\cot }^{-1}}\sqrt{\frac{41}{16}-1}\] \[\left( \because \,\,\text{cose}{{\text{c}}^{-1}}\,x={{\cot }^{-1}}\,\sqrt{{{x}^{2}}-1} \right)\] \[={{\cot }^{-1}}\,9+{{\cot }^{-1}}\frac{5}{4}\] \[={{\tan }^{-1}}\frac{1}{9}+{{\tan }^{-1}}\frac{4}{5}\] \[={{\tan }^{-1}}\left( \frac{\frac{1}{9}+\frac{4}{5}}{1-\frac{1}{9}.\frac{4}{5}} \right)\] \[={{\tan }^{-1}}\left( \frac{41}{41} \right)\] \[={{\tan }^{-1}}(1)=\frac{\pi }{4}\]
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