MCQ
If $\cos \theta=\frac{1}{2}\left(x+\frac{1}{x}\right)$, then $\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)=$
  • A
    $\sin 2 \theta$
  • $\cos 2 \theta$
  • C
    $\tan 2 \theta$
  • D
    $\sec 2 \theta$

Answer

Correct option: B.
$\cos 2 \theta$
(B)
Given that, $\cos \theta=\frac{1}{2}\left(x+\frac{1}{x}\right)$
$\Rightarrow x+\frac{1}{x}=2 \cos \theta$
Now, $x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2$
$=(2 \cos \theta)^2-2$
$=4 \cos ^2 \theta-2=2 \cos 2 \theta$
$\therefore \quad \frac{1}{2}\left(x^2+\frac{1}{x^2}\right)=\frac{1}{2} \times 2 \cos 2 \theta=\cos 2 \theta$

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