Question : If $7 \sin ^2 \theta+4 \cos ^2 \theta=5$ and $\theta$ lies in the first quadrant, then what is the value of $\frac{\sqrt{3} \sec \theta+\tan \theta}{\sqrt{2} \cot \theta-\sqrt{3} \cos \theta}$?

Option 1: $2(1+\sqrt{2})$

Option 2: $3 \sqrt{2}$

Option 3: $2(\sqrt{2}-1)$

Option 4: $4 \sqrt{2}$

Correct Answer: $2(1+\sqrt{2})$

Solution : $7 \sin ^2 \theta+4 \cos ^2 \theta=5$
$⇒7 \sin ^2 \theta+4 (1-\sin ^2 \theta)=5$
$⇒7 \sin ^2 \theta+4- 4\sin ^2 \theta=5$
$⇒3 \sin ^2 \theta=1$
$⇒\sin ^2 \theta=\frac{1}{3}$
$⇒\sin\theta=\frac{1}{\sqrt3}= \frac{P}{H}$
Now, by Pythagoras theorem
Where B is denoted as Base, H is denoted as Hypotenuse, and P is denoted as Perpendicular.
$B^2 = H^2 - P^2$
$⇒B^2 = (\sqrt{3})^2 - (1)^2$
$⇒B^2 = (3 -1)$
$⇒B^2 = 2$
$⇒B = \sqrt2$
Now,
The value of $\frac{\sqrt{3} \sec \theta+\tan \theta}{\sqrt{2} \cot \theta-\sqrt{3} \cos \theta}$
= $\frac{(\sqrt{3}\times \frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{\sqrt{2}})}{(\sqrt{2}\times\sqrt{2}-\sqrt{3}\times\frac{\sqrt{2}}{\sqrt{3}})}$
= $\frac{(\frac{3}{\sqrt{2}}+\frac{1}{\sqrt{2}})}{(2-\sqrt{2})}$
= $\frac{(\frac{4}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}})}{(2-\sqrt{2})}$
= $\frac{(\frac{4\sqrt{2}}{2})}{(2-\sqrt{2})}$
= $\frac{2\sqrt{2}}{(2-\sqrt{2})}$
Multiplying with conjugate ${(2+\sqrt{2})}$
= $\frac{2\sqrt{2}}{(2-\sqrt{2})}\times \frac{(2+\sqrt{2})}{(2+\sqrt{2})}$
= $\frac{2\sqrt{2}{(2+\sqrt{2})}}{2}$
= $\sqrt{2}{(2+\sqrt{2})}$
= $2\sqrt{2}+2$
= $2{(1+\sqrt{2})}$
Hence, the correct answer is $2{(1+\sqrt{2})}$.