frac{tan ^{2} theta}{sec theta+1}-sec theta i | vedclass

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Question

(A) We know that $1 + 	an^{2} 	heta = \sec^{2} 	heta$,which implies $	an^{2} 	heta = \sec^{2} 	heta - 1$.Substituting this into the expression:$

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$-1$

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(A) We know that $1 + 	an^{2} 	heta = \sec^{2} 	heta$,which implies $	an^{2} 	heta = \sec^{2} 	heta - 1$.Substituting this into the expression:$\frac{	an^{2} 	heta}{\sec 	heta + 1} - \sec 	heta = \frac{\sec^{2} 	heta - 1}{\sec 	heta + 1} - \sec 	heta$Using the algebraic identity $a^{2} - b^{2} = (a - b)(a + b)$,we can write $\sec^{2} 	heta - 1$ as $(\sec 	heta - 1)(\sec 	heta + 1)$:$= \frac{(\sec 	heta - 1)(\sec 	heta + 1)}{\sec 	heta + 1} - \sec 	heta$Canceling the common term $(\sec 	heta + 1)$:$= (\sec 	heta - 1) - \sec 	heta$$= \sec 	heta - 1 - \sec 	heta = -1$.

$\frac{\tan ^{2} \theta}{\sec \theta+1}-\sec \theta$ is equal to

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