SSC CGL 2022 PYQ — Question 2

The correct answer is option 4 i.e. 2(1+√2)

⇒ 7 sin² θ + 4 cos² θ = 5

⇒ 3 sin² θ + 4 sin² θ + 4 cos² θ = 5

⇒ 3 sin² θ = 1

⇒sinθ = 1/$\sqrt3$ = P/H

B² = 3 - 1 = 2

B = $\sqrt2$

⇒ secθ = $\sqrt{\frac{3}{2}}$ , tanθ = 1/$\sqrt2$ , cotθ = $\sqrt2$ and cosθ = $\sqrt{\frac{2}{3}}$

$\frac{\sqrt{3} sec \theta + tan \theta}{\sqrt{2} cot \theta - \sqrt{3} cos \theta}$ = $\frac{\sqrt3 \times \sqrt{\frac{3}{2}} + \frac{1}{\sqrt2}}{\sqrt2 \times \sqrt2 - \sqrt3 \times \sqrt{\frac{2}{3}}}$ = $\frac{2 \sqrt2}{2-\sqrt2}$

On further rationalization we get,

= $\frac{2 \sqrt2(2+\sqrt2)}{4-2}$ = 2(1+√2)