Question : If $2 \sec ^2 \theta-9 \sec \theta+7=0$, then what is the value of $\cos ^2 \theta+\sec ^2 \theta$?
Option 1: 2
Option 2: 3
Option 3: 5
Option 4: 1
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Correct Answer: 2
Solution :
Given: $2 \sec ^2 \theta-9 \sec \theta+7=0$
Let $\sec\theta = x$
Now,
$2x ^2-9x+7=0$
⇒ $2x ^2-9x+7=0$
⇒ $2x ^2-2x-7x+7=0$
⇒ $2x(x-1)-7(x-1)=0$
$\therefore x = 1, \frac{7}{2}$
Now, $\cos ^2 \theta+\sec ^2 \theta$
$=\frac{1}{\sec ^2 \theta}+\sec ^2 \theta$
$=\frac{1}{x^2}+x^2 $
Putting the value of $x=1$, we get:
$=1+1=2$
Hence, the correct answer is 2.
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