The geometric mean of tan 1^{circ}, tan 2^{ci | vedclass

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(B) Let $P = 	an 1^{\circ} \cdot 	an 2^{\circ} \cdot \ldots \cdot 	an 89^{\circ}$.Using the property $	an(90^{\circ} - 	heta) = \cot 	heta$,we c

Vedclass · Accepted Answer

$1$

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(B) Let $P = 	an 1^{\circ} \cdot 	an 2^{\circ} \cdot \ldots \cdot 	an 89^{\circ}$.Using the property $	an(90^{\circ} - 	heta) = \cot 	heta$,we can write the product as:$P = (	an 1^{\circ} \cdot 	an 89^{\circ}) \cdot (	an 2^{\circ} \cdot 	an 88^{\circ}) \cdot \ldots \cdot (	an 44^{\circ} \cdot 	an 46^{\circ}) \cdot 	an 45^{\circ}$.Since $	an 	heta \cdot 	an(90^{\circ} - 	heta) = 	an 	heta \cdot \cot 	heta = 1$,each pair equals $1$.Thus,$P = 1 \cdot 1 \cdot \ldots \cdot 1 \cdot 	an 45^{\circ} = 1 \cdot 1 = 1$.The geometric mean of $n$ terms $a_1, a_2, \ldots, a_n$ is $(a_1 \cdot a_2 \cdot \ldots \cdot a_n)^{1/n}$.Here,$n = 89$,so the geometric mean is $(P)^{1/89} = (1)^{1/89} = 1$.Hence,option $B$ is correct.

The geometric mean of $\tan 1^{\circ}, \tan 2^{\circ}, \ldots, \tan 89^{\circ}$ is

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