Simplify (1+tan ^{2} theta)(1-sin theta)(1+si | vedclass

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(A) Given expression: $(1+	an ^{2} 	heta)(1-\sin 	heta)(1+\sin 	heta)$Using the algebraic identity $(a-b)(a+b) = a^{2}-b^{2}$,we have $(1-\sin 	h

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(A) Given expression: $(1+	an ^{2} 	heta)(1-\sin 	heta)(1+\sin 	heta)$Using the algebraic identity $(a-b)(a+b) = a^{2}-b^{2}$,we have $(1-\sin 	heta)(1+\sin 	heta) = 1-\sin ^{2} 	heta$.So,the expression becomes $(1+	an ^{2} 	heta)(1-\sin ^{2} 	heta)$.We know that $1+	an ^{2} 	heta = \sec ^{2} 	heta$ and $1-\sin ^{2} 	heta = \cos ^{2} 	heta$.Substituting these,we get $\sec ^{2} 	heta \cdot \cos ^{2} 	heta$.Since $\sec 	heta = \frac{1}{\cos 	heta}$,we have $\sec ^{2} 	heta = \frac{1}{\cos ^{2} 	heta}$.Therefore,$\frac{1}{\cos ^{2} 	heta} \cdot \cos ^{2} 	heta = 1$.

Simplify $(1+\tan ^{2} \theta)(1-\sin \theta)(1+\sin \theta)$

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