SSC CGL 2024 PYQ — Question 70

The correct answer is option 4 i.e.  $\frac{9}{16}$.

$4\tan\theta - 3 = 0$  

$\tan\theta = \frac{3}{4}$

$\sin\theta = \frac{\tan\theta}{\sqrt{1 + \tan^2\theta}}$

$\cos\theta = \frac{1}{\sqrt{1 + \tan^2\theta}}$

⇒ $1 + \tan^2\theta = 1 + \left(\frac{3}{4}\right)^2 = 1 + \frac{9}{16} = \frac{25}{16}$

⇒ $\sin\theta = \frac{\frac{3}{4}}{\sqrt{\frac{25}{16}}} = \frac{\frac{3}{4}}{\frac{5}{4}} = \frac{3}{5}$

⇒ $\cos\theta = \frac{1}{\sqrt{\frac{25}{16}}} = \frac{1}{\frac{5}{4}} = \frac{4}{5}$

$\cos2\theta = 2\cos^2\theta - 1$

⇒ $\cos2\theta = 2 \times \left(\frac{4}{5}\right)^2 - 1 = 2 \times \frac{16}{25} - 1$

⇒ $\frac{32}{25} - 1 = \frac{32}{25} - \frac{25}{25} = \frac{7}{25}$

⇒ $\frac{1 - \cos2\theta}{1 + \cos2\theta} = \frac{1 - \frac{7}{25}}{1 + \frac{7}{25}}$

⇒ $\frac{\frac{25}{25} - \frac{7}{25}}{\frac{25}{25} + \frac{7}{25}}$

⇒ $\frac{\frac{18}{25}}{\frac{32}{25}} = \frac{18}{32} = \frac{9}{16}$