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Answered by
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Step-by-step explanation:
cot15
∘
.cot20
∘
.cot70
∘
.cot75
∘
is
=(cot15
∘
.cot75
∘
)(cot20
∘
cot70
∘
)
Now cot75
∘
=tan15
∘
and cot70
∘
=tan20
∘
cot(90−θ)=tanθ
Therefore, the given expression can be written as
(cot15
∘
tan15
∘
)(cot20
∘
tan20
Answered by
1
ahmm ahmm
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