যদি \(2x=secA\) এবং \(\cfrac{2}{x} =tanA\) হয়, তাহলে \(2(x^2-\cfrac{1}{x^2})\)-এর মান (a) \(\cfrac{1}{2}\) (b) \(\cfrac{1}{4}\) (c) \(\cfrac{1}{8}\) (d) \(\cfrac{1}{16}\)

Answer: A
\(2x+\cfrac{2}{x}=secA+tanA \)
বা, \(2(x+\cfrac{1}{x})=(secA+tanA)\)

আবার, \(2x-\cfrac{2}{x}=secA-tanA \)
বা, \(2(x-\cfrac{1}{x})=(secA-tanA)\)

\(∴2(x+\cfrac{1}{x})×2(x-\cfrac{1}{x})\)
\(=(secA+tanA)(secA-tanA) \)
বা, \(2×2(x^2-\cfrac{1}{x^2} )=sec^2⁡A-tan^2⁡A \)
বা, \(2×2(x^2-\cfrac{1}{x^2} )=1\)
বা, \(2(x^2-\cfrac{1}{x^2} )=\cfrac{1}{2}\)


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