Question

Prove that:

(sec8x - 1)÷(sec4x - 1) = tan8x/tan2x ?

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Answer 1

Answer:

(sec8x - 1)÷(sec4x - 1) = tan8x/tan2x

✔️Hope it will be helpful.✔️

Answer 2

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$\large{\bold{ \frac{sec8x-1}{sec4x-1}=\frac{tan8x}{tan2x}}}$

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LHS

$\frac{ \sec(8x) - 1 }{ \sec(4x) - 1 } \\ \\ = \frac{ \frac{1}{ \cos(8x) } - 1 }{ \frac{1}{ \cos(4x) } - 1 } \\ \\ = \frac{1 - \cos(8x) }{ \cos(8x) } \times \frac{ \cos(4x) }{1 - \cos(4x) } \\ \\ = \frac{2 { { \sin}^{2} (4x) } }{ \cos(8x) } \times \frac{ \cos(4x) }{2 {\sin}^{2}(2x) }$

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$\\ \\ since \: we \: know \: that \: \\ 1 - \cos(8x) = 2 { \sin}^{2} ( \frac{8x}{2} ) = 2 { \sin}^{2} (4x) \\ and \\ 1 - \cos(4x) = 2 { \sin}^{2}( \frac{4x}{2} ) = 2 { \sin}^{2} (2x) \\ \\ </p><p><strong> </strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong>_</strong><strong> </strong></p><p></p><p>[tex] \frac{2 \sin(4x) \cos(4x) }{ \cos(8x) } \times \frac{ \sin(4x) }{2 { \sin}^{2} (2x)} \\ \\ = ( \frac{2 \sin(4x) \cos(4x) }{ \cos(8x) } ) \times \frac{ \sin(4x) }{2 { \sin }^{2} (2x)} \\ \\ = ( \frac{2 \sin(4x) \cos(4x) }{ \cos(8x) } ) \times \frac{ \sin(4x) }{2 { \sin}^{2} (x)} \\ \\ ( \frac{2 \sin(4x) \cos(4x) }{ \cos(8x) } ) \times ( \frac{2 \sin(2x) \cos(2x) }{2 { \sin}^{2} (2x)} ) \\ \\ = ( \frac{ \sin2(4x) }{ \cos(8x) } ) \times \frac{ \cos(2x) }{ \sin(2x) } \\ \\ = ( \frac{ \sin(8x) }{ \cos(8x) } \times \frac{ \cos(2x) }{ \sin(2x) } \\ \\ = \tan(8x) \cot(2x) \\ \\ = \frac{ \tan(8x) }{ \tan(2x) }$

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