1
Question
tan75∘−cot75∘
[MNR 1982; Pb. CET 1990, 2000]
tan75∘−cot75∘
[MNR 1982; Pb. CET 1990, 2000]
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Solution
The correct option is A
tan75∘−cot75∘ = cot15∘−cot75∘
= (2+√3)−(2−√3)=2√3.
tan75∘−cot75∘ = cot15∘−cot75∘
= (2+√3)−(2−√3)=2√3.