Question : ABCD is a rectangle of which AC is a diagonal. The value of $(\tan^2\angle CAD+1)\sin^2\angle BAC$ is:
Option 1: $2$
Option 2: $\frac{1}{4}$
Option 3: $1$
Option 4: $0$
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Correct Answer: $1$
Solution :
Given: ABCD is a rectangle of which AC is a diagonal.
So, $\angle BAD = 90°$, then $\angle BAC = 90°- \theta$ and $\angle CAD = \theta$.
Also, $(\tan^2\angle CAD+1)\sin^2\angle BAC$
$=[\tan^{2} \ \theta + 1]\sin^{2}(90°-\theta)$
$=\sec^{2} \theta × \cos^{2} \theta$
$= 1$
Hence, the correct answer is $1$.
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