Question:
\( \cot^{-1}7+\cot^{-1}8+\cot^{-1}5= \)
\( \cot^{-1}7+\cot^{-1}8+\cot^{-1}5= \)
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Use \( \cot^{-1}x+\cot^{-1}y=\cot^{-1}\frac{xy-1}{x+y} \) twice.
Updated On: Nov 12, 2026
- \(\cot^{-1}2\)
- \(\cot^{-1}10\)
- \(\cot^{-1}6\)
- \(\cot^{-1}4\)
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The Correct Option is A
Solution and Explanation
\( \cot^{-1}7+\cot^{-1}8=\cot^{-1}\frac{56-1}{7+8}=\cot^{-1}\frac{55}{15}=\cot^{-1}\frac{11}{3} \).
Now \( \cot^{-1}\frac{11}{3}+\cot^{-1}5=\cot^{-1}\frac{\frac{11}{3}\cdot5-1}{\frac{11}{3}+5}=\cot^{-1}\frac{52/3}{26/3}=\cot^{-1}2 \).
Answer: (1) \( \cot^{-1}2 \)
Now \( \cot^{-1}\frac{11}{3}+\cot^{-1}5=\cot^{-1}\frac{\frac{11}{3}\cdot5-1}{\frac{11}{3}+5}=\cot^{-1}\frac{52/3}{26/3}=\cot^{-1}2 \).
Answer: (1) \( \cot^{-1}2 \)
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