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: $\frac{2-\sqrt{3}}{2}$
Option 2: $\frac{1+2 \sqrt{3}}{2}$
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: $4-2 \sin ^2 \theta-5 \cos \theta=0$
⇒ $4-2(1- \cos ^2 \theta)-5 \cos \theta=0$
⇒ $4-2+ 2\cos ^2 \theta-5 \cos \theta=0$
⇒ $2\cos ^2 \theta-5 \cos \theta+2=0$
⇒ $2\cos ^2 \theta-4 \cos \theta- \cos \theta+2=0$
⇒ $2\cos \theta(\cos \theta-2) - 1(\cos \theta-2) = 0$
⇒ $(2\cos \theta-1)(\cos \theta-2) = 0$
⇒ $\cos \theta = \frac{1}{2}$ and $\cos \theta = 2$
Rejecting $\cos \theta = 2$ as $0^{\circ}<\theta<90^{\circ}$
⇒ $\cos \theta = \frac{1}{2}$
⇒ $\theta = 60^{\circ}$
So, $\cos \theta + \tan \theta$
$= \cos 60^{\circ} + \tan 60^{\circ}$
$ = \frac{1}{2} + \sqrt3$
$= \frac{1+2\sqrt3}{2}$
Hence, the correct answer is $\frac{1+2\sqrt3}{2}$.
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