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$. Thus,$	an 89^\circ = 	an(90^\circ - 1^\circ) = \cot 1^\circ$. Similarly,$	an 88^\circ = \

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(A) We know that $	an(90^\circ - 	heta) = \cot 	heta$. Thus,$	an 89^\circ = 	an(90^\circ - 1^\circ) = \cot 1^\circ$. Similarly,$	an 88^\circ = \cot 2^\circ$,and so on. The expression is $P = 	an 1^\circ 	an 2^\circ \dots 	an 44^\circ 	an 45^\circ 	an 46^\circ \dots 	an 88^\circ 	an 89^\circ$. Pairing terms from both ends: $(	an 1^\circ 	an 89^\circ) = (	an 1^\circ \cot 1^\circ) = 1$. $(	an 2^\circ 	an 88^\circ) = (	an 2^\circ \cot 2^\circ) = 1$. Continuing this up to $	an 44^\circ 	an 46^\circ = 1$. The middle term is $	an 45^\circ = 1$. Therefore,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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