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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