Question : If $\sec\theta-\cos\theta=\frac{3}{2}$, where $\theta$ is a positive acute angle, then the value of $\sec\theta$ is:
Option 1: $-\frac{1}{2}$
Option 2: $1$
Option 3: $2$
Option 4: $0$
Correct Answer: $2$
Solution :
Given: $\sec\theta-\cos\theta$ = $\frac{3}{2}$
⇒ $\frac{1}{\cos\theta}-\cos\theta=\frac{3}{2}$
Let $\cos\theta=x$
⇒ $\frac{1}{x}-x=\frac{3}{2}$
⇒ $2(1-x^2) = 3x$
⇒ $2-2x^2 = 3x$
⇒ $2x^2+3x - 2=0$
⇒ $2x^2+4x-x-2 = 0$
⇒ $2x(x+2)-1(x+2) = 0$
⇒ $(x+2)(2x-1) = 0$
⇒ $x = -2, \frac{1}{2}$
Neglecting the negative value, we get,
$\cos\theta = \frac{1}{2}$
So, $\sec\theta = 2$
Hence, the correct answer is $2$.
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