MCQ
If $2 x=\sec \theta$ and $\frac{2}{x}=\tan \theta$, then $2\left(x^2-\frac{1}{x^2}\right)=$
- ✓$\frac{1}{2}$
- B$2$
- C$\frac{1}{4}$
- D$4$
✓
Answer
Correct option: A.
$\frac{1}{2}$
We have, $2 x=\operatorname{sect}$ and $\frac{2}{x}=\tan \theta$
$\Rightarrow x=\frac{\sec \theta}{2}$ and $\frac{1}{x}=\frac{\tan \theta}{2}$
$\therefore 2\left(x^2-\frac{1}{x^2}\right)-2\left(\frac{\sec ^2 \theta}{4}-\frac{\tan ^2 \theta}{4}\right) \quad\left[\because \sec ^2 \theta-\tan ^2 \theta=1\right]$
$=\frac{2}{4}\left(\sec ^2 \theta-\tan ^2 \theta\right)=\frac{1}{2} \cdot 1 \quad[(\theta)]$
$=\frac{1}{2}$
$\Rightarrow x=\frac{\sec \theta}{2}$ and $\frac{1}{x}=\frac{\tan \theta}{2}$
$\therefore 2\left(x^2-\frac{1}{x^2}\right)-2\left(\frac{\sec ^2 \theta}{4}-\frac{\tan ^2 \theta}{4}\right) \quad\left[\because \sec ^2 \theta-\tan ^2 \theta=1\right]$
$=\frac{2}{4}\left(\sec ^2 \theta-\tan ^2 \theta\right)=\frac{1}{2} \cdot 1 \quad[(\theta)]$
$=\frac{1}{2}$
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