Question : If $\theta$ is a positive acute angle and $4\cos ^{2}\theta -4\cos \theta +1=0$, then the value of $\tan (\theta -15^{\circ})$is equal to:
Option 1: $0$
Option 2: $1$
Option 3: $\sqrt{3}$
Option 4: $\frac{1}{\sqrt{3}}$
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Correct Answer: $1$
Solution :
Given: $4\cos ^{2}\theta -4\cos \theta +1=0$
The above equation is written as
$⇒(2\cos \theta - 1)^2 = 0$
$⇒2\cos \theta-1 = 0$
$⇒\cos \theta = \frac{1}{2}$
$⇒\cos \theta = \cos60^{\circ}$
$⇒\theta = 60^{\circ}$
Now, putting the value of $\theta$ in $\tan (\theta -15^{\circ})$, we get,
$\tan (60^{\circ}-15^{\circ})=\tan (45^{\circ}) = 1$
Hence, the correct answer is $1$.
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