SSC CGL 2020 PYQ — Question 65

The correct answer is option 2 i.e.   $\frac{{\sqrt {{b^2} - {a^2}} }}{a}$

cosec θ = b/a = hypotenuse/perpendicular

(base) = $\sqrt{b^2 - a^2}$

Cot θ = base/perpendicular = $\sqrt{b^2 - a^2}$/a

Tan θ = perpendicular/base = a/$\sqrt{b^2 - a^2}$.

⇒$\frac{\sqrt{3}cot\theta+1}{tan\theta+\sqrt{}3}$ 

= $\frac{\sqrt3 \times Cot \theta + 1}{\frac{1}{cot \theta} + \sqrt3}$ 

= Cot $\theta$ =   $\sqrt{b^2 - a^2}$/a.