Question
Short-Answer Questions:If $2\text{x}=\sec\text{A}$ and $\frac{2}{\text{x}}=\tan\text{A},$ prove that $\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=\frac{1}{4}.$
✓
Answer
Given: $\text{2x}=\sec\text{A}$$\Rightarrow\text{x}=\frac{\sec\text{A}}{2}\dots(\text{i})$
and $\frac{2}{\text{x}}=\tan\text{A}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\tan\text{A}}{2}\dots(\text{ii})$
$\therefore\text{x}+\frac{1}{\text{x}}=\frac{\sec\text{A}}{2}+\frac{\tan\text{A}}{2}$ $[\because$ From (i) and (ii)$]$
Also, $\text{x}-\frac{1}{\text{x}}=\frac{\sec\text{A}}{2}-\frac{\tan\text{A}}{2}$
$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)=\Big(\frac{\sec\text{A}}{2}+\frac{\tan\text{A}}{2}\Big)\Big(\frac{\sec\text{A}}{2}-\frac{\tan\text{A}}{2}\Big)$
$\Rightarrow\text{x}^2-\frac{1}{\text{x}^2}=\frac{1}{4}\big(\sec^2\text{A}-\tan^2\text{A}\big)$
$\therefore\ \text{x}^2-\frac{1}{\text{x}^2}=\frac{1}{4}\times1\ \big(\because\sec^2\text{A}-\tan^2\text{A}=1\big)$
$=\frac{1}{4}$
Hence proved.
and $\frac{2}{\text{x}}=\tan\text{A}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\tan\text{A}}{2}\dots(\text{ii})$
$\therefore\text{x}+\frac{1}{\text{x}}=\frac{\sec\text{A}}{2}+\frac{\tan\text{A}}{2}$ $[\because$ From (i) and (ii)$]$
Also, $\text{x}-\frac{1}{\text{x}}=\frac{\sec\text{A}}{2}-\frac{\tan\text{A}}{2}$
$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)=\Big(\frac{\sec\text{A}}{2}+\frac{\tan\text{A}}{2}\Big)\Big(\frac{\sec\text{A}}{2}-\frac{\tan\text{A}}{2}\Big)$
$\Rightarrow\text{x}^2-\frac{1}{\text{x}^2}=\frac{1}{4}\big(\sec^2\text{A}-\tan^2\text{A}\big)$
$\therefore\ \text{x}^2-\frac{1}{\text{x}^2}=\frac{1}{4}\times1\ \big(\because\sec^2\text{A}-\tan^2\text{A}=1\big)$
$=\frac{1}{4}$
Hence proved.
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