If sec ^{2} theta+tan ^{2} theta=frac{13}{12} | vedclass

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(B) We know the algebraic identity $a^{2}-b^{2}=(a-b)(a+b)$.Applying this to the expression $\sec ^{4} 	heta-	an ^{4} 	heta$,we get:$\sec ^{4} 	he

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$\frac{13}{12}$

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(B) We know the algebraic identity $a^{2}-b^{2}=(a-b)(a+b)$.Applying this to the expression $\sec ^{4} 	heta-	an ^{4} 	heta$,we get:$\sec ^{4} 	heta-	an ^{4} 	heta = (\sec ^{2} 	heta - 	an ^{2} 	heta)(\sec ^{2} 	heta + 	an ^{2} 	heta)$.From the fundamental trigonometric identity,we know that $\sec ^{2} 	heta - 	an ^{2} 	heta = 1$.Substituting the given values:$\sec ^{4} 	heta - 	an ^{4} 	heta = (1) 	imes \left(\frac{13}{12}ight) = \frac{13}{12}$.

If $\sec ^{2} \theta+\tan ^{2} \theta=\frac{13}{12}$,then the value of $\sec ^{4} \theta-\tan ^{4} \theta$ is .........

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