Question : If $\sqrt{2} \sec ^2 \theta-4 \sec \theta+2 \sqrt{2}=0$, then what is the value $\sin ^2 \theta+\tan ^2 \theta$?
Option 1: $\frac{1}{2}$
Option 2: $\frac{2}{3}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{3}{2}$
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Correct Answer: $\frac{3}{2}$
Solution :
$\sqrt{2} \sec ^2 \theta-4 \sec \theta+2 \sqrt{2}=0$
$⇒\sqrt{2} \sec ^2 \theta-2 \sec \theta-2 \sec \theta+2 \sqrt{2}=0$
$⇒\sqrt{2} \sec \theta( \sec \theta-\sqrt2)-2( \sec \theta- \sqrt{2})=0$
$⇒( \sec \theta-\sqrt2)(\sqrt{2} \sec \theta-2)=0$
$⇒( \sec \theta-\sqrt2)=0$
$⇒ \sec \theta=\sqrt2=\sec 45^\circ$
$⇒ \theta=45^\circ$
$\therefore\sin ^2 \theta+\tan ^2 \theta=\sin ^2 45^\circ+\tan ^245^\circ=\frac{1}{2}+1=\frac{3}{2}$
Hence, the correct answer is $\frac{3}{2}$.
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