(cot−1x)2−7(cot−1x)+10>0 (cot−1x−5)(cot−1−2)>0 cot−1x∈(−∞,2)∪(5,∞) But cot−1x lies in (0,π) Now, from equation (1) cot−1x∈(0,2) Now, it is clear from the graph x∈(cot2,∞)
All x satisfying the inequality (cot−1x)2−7(cot−1x)+10> 0 , lie in the interval :
Held on 11 Jan 2019 · Verified 6 Jul 2026.
(−∞,cot5)∪(cot4,cot2)
(cot2,∞)
(−∞,cot5)∪(cot2,∞)
(cot5,cot4)
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