SSC CGL 2023 PYQ — Question 66

The correct answer is option 1 i.e. 2cos θ.

cos2θ = cos²θ − sin²θ = 2cos²θ − 1

⇒ 2cos²θ = 1 + cos2θ

According to the question¸

$\sqrt{2 + \sqrt{2+\sqrt{2+2 \cos 8 \theta}}}$

⇒ $\sqrt{2 + \sqrt{2+\sqrt{2+(1 \cos 8 \theta)}}}$

⇒ $\sqrt{2 + \sqrt{2+2 \cos 4 \theta}}$

⇒ $\sqrt{2 + \sqrt{2(1+ \cos 4 \theta)}}$

⇒ $\sqrt{2 + 2 \cos 2 \theta)}$

⇒ $\sqrt{2 (1+ \cos 2 \theta)}$

⇒ 2cosθ = RHS