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: $4 \sqrt{2}$
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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 $2{(1+\sqrt{2})}$.
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