If (sec A + tan A)(sec B + tan B)(sec C + tan | vedclass

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(D) Let $L = (\sec A + 	an A)(\sec B + 	an B)(\sec C + 	an C)$ and $M = (\sec A - 	an A)(\sec B - 	an B)(\sec C - 	an C)$.Given that $L = M$.We

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$1$ or $-1$

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(D) Let $L = (\sec A + 	an A)(\sec B + 	an B)(\sec C + 	an C)$ and $M = (\sec A - 	an A)(\sec B - 	an B)(\sec C - 	an C)$.Given that $L = M$.We know that $(\sec 	heta + 	an 	heta)(\sec 	heta - 	an 	heta) = \sec^2 	heta - 	an^2 	heta = 1$.Multiplying $L$ and $M$,we get $LM = (\sec^2 A - 	an^2 A)(\sec^2 B - 	an^2 B)(\sec^2 C - 	an^2 C) = 1 	imes 1 	imes 1 = 1$.Since $L = M$,we can substitute $M$ with $L$ in the equation $LM = 1$,which gives $L^2 = 1$.This implies $L = 1$ or $L = -1$.Since $L = M$,both sides are equal to $1$ or $-1$.

If $(\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C),$ then each side is equal to

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