tan 23^{circ} tan 42^{circ} tan 48^{circ} tan | vedclass

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(D) We know that $	an 	heta = \cot(90^{\circ} - 	heta)$.Using this identity:$	an 48^{\circ} = 	an(90^{\circ} - 42^{\circ}) = \cot 42^{\circ}$$	a

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(D) We know that $	an 	heta = \cot(90^{\circ} - 	heta)$.Using this identity:$	an 48^{\circ} = 	an(90^{\circ} - 42^{\circ}) = \cot 42^{\circ}$$	an 67^{\circ} = 	an(90^{\circ} - 23^{\circ}) = \cot 23^{\circ}$Now,substitute these values into the expression:$	an 23^{\circ} 	an 42^{\circ} 	an 48^{\circ} 	an 67^{\circ} = 	an 23^{\circ} 	an 42^{\circ} \cot 42^{\circ} \cot 23^{\circ}$Rearranging the terms:$= (	an 23^{\circ} \cot 23^{\circ}) 	imes (	an 42^{\circ} \cot 42^{\circ})$Since $	an 	heta 	imes \cot 	heta = 1$:$= 1 	imes 1 = 1$

$\tan 23^{\circ} \tan 42^{\circ} \tan 48^{\circ} \tan 67^{\circ} = \ldots \ldots \ldots \ldots .$

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