Question : If $\frac{\sin ^2 \theta}{\cos ^2 \theta-3 \cos \theta+2}=1, \theta$ lies in the first quadrant, then the value of $\frac{\tan ^2 \frac{\theta}{2}+\sin ^2 \frac{\theta}{2}}{\tan \theta+\sin \theta}$ is:

Option 1: $\frac{2 \sqrt{3}}{27}$

Option 2: $\frac{5 \sqrt{3}}{27}$

Option 3: $\frac{2 \sqrt{3}}{9}$

Option 4: $\frac{7 \sqrt{3}}{54}$

Correct Answer: $\frac{7 \sqrt{3}}{54}$

Solution : Given that $\frac{\sin ^2 \theta}{\cos ^2 \theta-3 \cos \theta+2}=1$,
$⇒\sin ^2 \theta = \cos ^2 \theta-3 \cos \theta+2$
$⇒\cos ^2 \theta - \sin ^2 \theta - 3 \cos \theta + 2 = 0$
Since $\cos ^2 \theta = 1 - \sin ^2 \theta$,
$⇒1 - 2 \sin ^2 \theta - 3 \cos \theta + 2 = 0$
$⇒2 \sin ^2 \theta + 3 \cos \theta - 3 = 0$
Now, we know that $\sin ^2 \theta = 1 - \cos ^2 \theta$,
$⇒2 (1 - \cos ^2 \theta) + 3 \cos \theta - 3 = 0$
$⇒2 - 2 \cos ^2 \theta + 3 \cos \theta - 3 = 0$
$⇒2 \cos ^2 \theta - 3 \cos \theta + 1 = 0$
$⇒( \cos \theta -1)( 2\cos \theta-1 )=0$
$⇒\cos \theta = \frac{1}{2}$ or,
$⇒\cos \theta = 1$
$⇒\theta=60^{\circ}$ or $\theta=0^{\circ}$
But $\theta=0^{\circ}$ does not satisfy the given equation,
So, $\theta =60^{\circ}$
Substituting these values into the expression,
$\frac{\tan ^2 \frac{\theta}{2}+\sin ^2 \frac{\theta}{2}}{\tan \theta+\sin \theta}=\frac{\tan ^2 \frac{60^{\circ}}{2}+\sin ^2 \frac{60^{\circ}}{2}}{\tan60^{\circ}+\sin 60^{\circ}}=\frac{\tan ^230^{\circ}+\sin ^230^{\circ}}{\tan60^{\circ}+\sin 60^{\circ}}=\frac{\frac{1}{3}+ \frac{1}{4}}{\sqrt3+ \frac{\sqrt3}{2}}=\frac{\frac{7}{12}}{ \frac{3\sqrt3}{2}}=\frac{7 \sqrt{3}}{54}$
Hence, the correct answer is $\frac{7 \sqrt{3}}{54}$.