SSC Selection Post 2019 PYQ — Question 71

The correct answer is Option 1 i.e. $\frac{2\sqrt3 +1}{2}$

$1 - 3(\cot^2\phi - \cos^2\phi) = \sin^2\phi$

$\cot^2\phi = \frac{1}{\sin^2\phi} - 1,\quad \cos^2\phi = 1 - \sin^2\phi$

$1 - 3\left(\frac{1}{\sin^2\phi} - 1 - (1 - \sin^2\phi)\right) = \sin^2\phi \Rightarrow 1 - 3\left(\frac{1}{\sin^2\phi} + \sin^2\phi - 2\right) = \sin^2\phi$

Let x = $\sin^2\phi$

$1 - 3\left(\frac{1}{x} + x - 2\right) = x \Rightarrow 4x^2 - 7x + 3 = 0 \Rightarrow x = \frac{3}{4} \Rightarrow \sin\phi = \frac{\sqrt{3}}{2}, \quad \cos\phi = \frac{1}{2}$

$2\sin\phi + \cos\phi = 2 \times \frac{\sqrt{3}}{2} + \frac{1}{2} = \sqrt{3} + \frac{1}{2}$