tan 1^circ tan 2^circ tan 3^circ tan 4^circ d | vedclass

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Question

(A) We know that $	an(90^\circ - 	heta) = \cot 	heta$.Therefore,$	an 89^\circ = 	an(90^\circ - 1^\circ) = \cot 1^\circ$.Similarly,$	an 88^\circ

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$1$

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(A) We know that $	an(90^\circ - 	heta) = \cot 	heta$.Therefore,$	an 89^\circ = 	an(90^\circ - 1^\circ) = \cot 1^\circ$.Similarly,$	an 88^\circ = \cot 2^\circ$,and so on.The expression can be written as: $(	an 1^\circ \cdot 	an 89^\circ) \cdot (	an 2^\circ \cdot 	an 88^\circ) \dots (	an 44^\circ \cdot 	an 46^\circ) \cdot 	an 45^\circ$.Since $	an 	heta \cdot \cot 	heta = 1$,each pair $(	an 	heta \cdot 	an(90^\circ - 	heta)) = 1$.There are $44$ such pairs,and $	an 45^\circ = 1$.Thus,the product is $1 	imes 1 	imes \dots 	imes 1 = 1$.

$\tan 1^\circ \tan 2^\circ \tan 3^\circ \tan 4^\circ \dots \tan 89^\circ = $

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