A) \( \Large \sin 2 \theta \) |
B) \( \Large \sec 2 \theta \) |
C) \( \Large \tan 2 \theta \) |
D) \( \Large \cos 2 \theta \) |
D) \( \Large \cos 2 \theta \) |
Given that \( \Large \cos \theta = \frac{1}{2} \left(x+\frac{1}{x}\right) => x+\frac{1}{x}=2 \cos \theta \)
We know that \( \Large x^{2}+\frac{1}{x^{2}} = \left(x+\frac{1}{x}\right)^{2}-2 \)
= \( \Large \left(2 \cos \theta \right)^{2}-2=4 \cos^{2} \theta -2 \)
= \( \Large 2 \cos 2 \theta \) .. From (i)
Therefore, \( \Large \frac{1}{2} \left(x^{2}+\frac{1}{x^{2}}\right)=\frac{1}{2} \times 2 \cos \theta = \cos 2 \theta \)