tan 6^{circ} tan 42^{circ} tan 66^{circ} tan | vedclass

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(B) We use the identity: $	an (60^{\circ}-A) 	an A 	an (60^{\circ}+A) = 	an 3A$. Given expression: $E = 	an 6^{\circ} 	an 42^{\circ} 	an 66^{\c

Vedclass · Accepted Answer

$1$

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(B) We use the identity: $	an (60^{\circ}-A) 	an A 	an (60^{\circ}+A) = 	an 3A$. Given expression: $E = 	an 6^{\circ} 	an 42^{\circ} 	an 66^{\circ} 	an 78^{\circ}$. Rearranging the terms: $E = (	an 6^{\circ} 	an 66^{\circ}) 	imes (	an 42^{\circ} 	an 78^{\circ})$. Using the identity $	an (60^{\circ}-A) 	an (60^{\circ}+A) = \frac{	an 3A}{	an A}$: For the first part: $	an 6^{\circ} 	an 66^{\circ} = 	an 6^{\circ} 	an (60^{\circ}+6^{\circ}) = \frac{	an 18^{\circ}}{	an 54^{\circ}}$. For the second part: $	an 42^{\circ} 	an 78^{\circ} = 	an (60^{\circ}-18^{\circ}) 	an (60^{\circ}+18^{\circ}) = \frac{	an 54^{\circ}}{	an 18^{\circ}}$. Multiplying these: $E = \left( \frac{	an 18^{\circ}}{	an 54^{\circ}} ight) 	imes \left( \frac{	an 54^{\circ}}{	an 18^{\circ}} ight) = 1$.

$\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ} = $

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