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\(cot^{-1}9+cosec^{-1}\frac{\sqrt{41}}{4}\)= ?A. \(\frac{\pi}{6}\)B. \(\frac{\pi}{4}\)C. \(\frac{\pi}{3}\)D. \(\frac{3\pi}{4}\) |
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Answer» Correct Answer is (B) \(\frac{\pi}{4}\) Now \(cot^{-1}9+cosec^{-1}\frac{\sqrt{41}}{4}\) can be written in terms of tan inverse as \(cot^{-1}9+cosec^{-1}\frac{\sqrt{41}}{4}\)= \(cot^{-1}\frac{1}{9}+cosec^{-1}\frac{4}{5}\) Since we know that tan-1 x + tan-1 y = \(tan^{-1}(\frac{x+y}{1-xy})\) ⇒ \(cos^{-1}\frac{1}{9}+cosec^{-1}\frac{4}{5}\) = tan-1\((\frac{\frac{1}{9}+\frac{4}{5}}{1-(\frac{1}{9}\times\frac{4}{5})})\) = tan-1\((\frac{41}{41})\) = tan-1 (1) = \(\frac{\pi}{4}\) |
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