If x-1x-2cos- then what is the value of x2-1x2-

The correct option is C 2cos ^ 2 α 2sin ^ 2 α x^2+1x^2 = ( x+1x )^2 2⇒ x^2+1x^2 = 4cos^2α 2 ⇒ 4cos^2 2(sin^2α +cos^2α ) = 2cos^2α 2sin^2α ...

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Circular Measurement of Angle

If x+1x=2co...

Question

If $x + \frac{1}{x} = 2 cos α$ , then what is the value of $x^{2} + \frac{1}{x^{2}}$ ?

4 {cos}^{2} α

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4 {cos}^{2} α - 1

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2 {cos}^{2} α - 2 {sin}^{2} α

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{cos}^{2} α - {sin}^{2} α

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Solution

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(\frac{1}{{sec}^{2} α - {cos}^{2} α} + \frac{1}{csc^{2} α - {sin}^{2} α}) \times {sin}^{2} α {cos}^{2} α = \frac{1 - {sin}^{2} α {cos}^{2} α}{2 + {sin}^{2} α {cos}^{2} α}

(\frac{1}{{sec}^{2} α - {cos}^{2} α} + \frac{1}{{csc}^{2} α - {sin}^{2} α}) {cos}^{2} α . {sin}^{2} α = \frac{1 - {cos}^{2} α . {sin}^{2} α}{2 + {cos}^{2} α . {sin}^{2} α}

∣ ∣ ∣ \begin{matrix} {sin}^{2} α & {cos}^{2} α {cos}^{2} α & {sin}^{2} α \end{matrix} ∣ ∣ ∣ = 0, α ϵ (0, π)

α

α

(0 < α < π / 2)

\frac{x^{2}}{{cos}^{2} α} - \frac{y^{2}}{{sin}^{2} α} = 1

ϵ

α = \frac{x^{2}}{(1 + x^{4})}

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