Question:
\(\cot^{-1}(1) + \cot^{-1}(2) + \cot^{-1}(3) =\)
\(\cot^{-1}(1) + \cot^{-1}(2) + \cot^{-1}(3) =\)
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\(\cot^{-1}(2) + \cot^{-1}(3) = \frac{\pi}{4}\).
Updated On: Apr 27, 2026
- \(\frac{\pi}{4}\)
- \(\frac{\pi}{2}\)
- \(\frac{3\pi}{2}\)
- \(\pi\)
- 0
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Use \(\cot^{-1} x = \tan^{-1} \frac{1}{x}\) and formula \(\tan^{-1} a + \tan^{-1} b = \tan^{-1} \frac{a+b}{1-ab}\).
Step 2: Detailed Explanation:
\(\cot^{-1}(1) = \tan^{-1}(1) = \frac{\pi}{4}\)
\(\cot^{-1}(2) + \cot^{-1}(3) = \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{1 - \frac{1}{6}}\right) = \tan^{-1}\left(\frac{5/6}{5/6}\right) = \tan^{-1}(1) = \frac{\pi}{4}\)
Sum = \(\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}\)
Step 3: Final Answer:
\(\frac{\pi}{2}\).
Use \(\cot^{-1} x = \tan^{-1} \frac{1}{x}\) and formula \(\tan^{-1} a + \tan^{-1} b = \tan^{-1} \frac{a+b}{1-ab}\).
Step 2: Detailed Explanation:
\(\cot^{-1}(1) = \tan^{-1}(1) = \frac{\pi}{4}\)
\(\cot^{-1}(2) + \cot^{-1}(3) = \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{1 - \frac{1}{6}}\right) = \tan^{-1}\left(\frac{5/6}{5/6}\right) = \tan^{-1}(1) = \frac{\pi}{4}\)
Sum = \(\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}\)
Step 3: Final Answer:
\(\frac{\pi}{2}\).
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