Question

If secA=2x and tanA=2/x, find the value of 2(x^2-1/x^2)

Answer 1

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Answer 2

The value of 2($x^{2}-\dfrac{1}{x^{2} }$) is $\dfrac{1}{2}$.

Step-by-step explanation:

We have,

$secA=2x$ and $\tan A=\dfrac{2}{x}$

To find,  the value of $2(x^2-\dfrac{1}{x^{2}} )=?$

∴$\sec ^{2} A=4x^{2}$               ...(1)

and

$\tan ^{2}  A=\dfrac{4}{x^{2} }$   ...(2)

Subtracting (1) from (2), we get

$4x^{2}-\dfrac{4}{x^{2} }=\sec^{2} A-\tan^{2} A$

⇒$4(x^{2}-\dfrac{1}{x^{2} })=1$

[ ∵ $\sec^{2} x-\tan^{2} x =1$]

⇒ 2($x^{2}-\dfrac{1}{x^{2} })=\dfrac{1}{2}$

Hence, the value of 2($x^{2}-\dfrac{1}{x^{2} }$) is $\dfrac{1}{2}$.