ABCD is a rectangle where AC is a diagonal. T | vedclass

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(C) In a rectangle $ABCD$,the diagonal $AC$ divides the rectangle into two congruent right-angled triangles $	riangle ABC$ and $	riangle ADC$.Let $\

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$1$

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(C) In a rectangle $ABCD$,the diagonal $AC$ divides the rectangle into two congruent right-angled triangles $	riangle ABC$ and $	riangle ADC$.Let $\angle CAD = 	heta$. Since $AD \parallel BC$,$\angle CAD = \angle ACB = 	heta$.Also,$\angle BAC = 90^{\circ} - 	heta$.Using the identity $1 + 	an^2 	heta = \sec^2 	heta$,the expression becomes:$(	an^2 \angle CAD + 1) \sin^2 \angle BAC = \sec^2 \angle CAD \cdot \sin^2 \angle BAC$$= \frac{1}{\cos^2 \angle CAD} \cdot \sin^2 (90^{\circ} - \angle CAD)$$= \frac{1}{\cos^2 \angle CAD} \cdot \cos^2 \angle CAD = 1$.

$ABCD$ is a rectangle where $AC$ is a diagonal. The value of $(\tan^2 \angle CAD + 1) \sin^2 \angle BAC$ is

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