Question : If $4-2 \sin ^2 \theta-5 \cos \theta=0,0^{\circ}<\theta<90^{\circ}$, then the value of $\cos \theta-\tan \theta$ is:
Option 1: $\frac{1+2 \sqrt{3}}{2}$
Option 2: $\frac{2-\sqrt{3}}{2}$
Option 3: $\frac{2+\sqrt{3}}{2}$
Option 4: $\frac{1-2 \sqrt{3}}{2}$
Correct Answer: $\frac{1-2 \sqrt{3}}{2}$
Solution :
Given: $4 - 2\sin^2 \theta - 5\cos \theta = 0$
⇒ $4 - 2 × (1 - \cos^2\theta) - 5cos\theta = 0$
⇒ $4 - 2 + 2\cos^2\theta - 5\cos\theta = 0$
⇒ $2\cos^2\theta - 5\cos\theta + 2 = 0$
⇒ $2\cos^2\theta - 4\cos\theta - \cos\theta + 2 = 0$
⇒ $2\cos\theta × (\cos\theta - 2) - 1 × (\cos\theta - 2) = 0$
⇒ $(2\cos\theta - 1) × (\cos\theta - 2) = 0$
⇒ $\cos\theta = \frac{1}{2}, 2$
Rejecting $\cos\theta = 2$ as $0^\circ < \theta < 90^\circ$
So, $\cos\theta=\frac{1}{2}$
⇒ $\theta = 60^\circ$
The value of $\cos \theta -\tan \theta = \cos60^\circ - \tan60^\circ$
= $\frac{1}{2} -\sqrt3$
= $\frac{1-2 \sqrt{3}}{2}$
Hence, the correct answer is $\frac{1-2 \sqrt{3}}{2}$.
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