MCQ
If $2 x=\sec A$ and $\frac{2}{x}=\tan A$ then $2\left(x^2-\frac{1}{x^2}\right)=$ ?
- ✓$\frac{1}{2}$
- B$\frac{1}{4}$
- C$\frac{1}{16}$
- D$\frac{1}{8}$
✓
Answer
Correct option: A.
$\frac{1}{2}$
We know that $\sec ^2 A-\tan ^2 A=1$
$\therefore(2 x)^2-\left(\frac{2}{x}\right)^2=1 $
$\Rightarrow 4 x^2-\frac{4}{x^2}=1 $
$\Rightarrow 4\left(x^2-\frac{1}{x^2}\right)=1$
$\Rightarrow\left(x^2-\frac{1}{x^2}\right)=\frac{1}{4} $
$\Rightarrow 2\left(x^2-\frac{1}{x^2}\right)$
$=2 \times \frac{1}{4}$
$=\frac{1}{2}$
$\therefore(2 x)^2-\left(\frac{2}{x}\right)^2=1 $
$\Rightarrow 4 x^2-\frac{4}{x^2}=1 $
$\Rightarrow 4\left(x^2-\frac{1}{x^2}\right)=1$
$\Rightarrow\left(x^2-\frac{1}{x^2}\right)=\frac{1}{4} $
$\Rightarrow 2\left(x^2-\frac{1}{x^2}\right)$
$=2 \times \frac{1}{4}$
$=\frac{1}{2}$
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