Question

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Answer 1

Required Answer:-

Given to Prove:

$\rm \mapsto \dfrac{1 - { \sin}^{4}(x) }{ \cos^{4} (x) } = 2 { \sec}^{2} (x) - 1$

Proof:

Taking LHS,

$\rm \dfrac{1 - { \sin}^{4}(x) }{ \cos^{4} (x) }$

$\rm = \dfrac{ {(1)}^{2} - {( \sin}^{2}(x))^{2} }{ \cos^{4} (x) }$

$\rm = \dfrac{(1 + { \sin }^{2}(x))(1 - { \sin}^{2}(x)) }{ \cos^{4} (x) }$

We know that,

$\rm { \cos}^{2}(x) = 1 - { \sin}^{2} (x)$

Therefore,

$\rm \dfrac{(1 + { \sin }^{2}(x))(1 - { \sin}^{2}(x)) }{ \cos^{4} (x) }$

$\rm = \dfrac{(1 + { \sin }^{2}(x)) { \cos}^{2}(x) }{ \cos^{4} (x) }$

$\rm = \dfrac{1 + { \sin }^{2}(x) }{ \cos^{2} (x) }$

$\rm = \dfrac{1}{ \cos^{2} (x) } + \dfrac{ { \sin}^{2}(x) }{ { \cos}^{2}(x) }$

$\rm = { \sec}^{2}(x) + { \tan}^{2} (x)$

We know that,

$\rm { \sec}^{2}(x) - { \tan}^{2} (x) = 1 \implies { \tan}^{2}(x) = { \sec}^{2} (x) - 1$

Therefore,

$\rm { \sec}^{2}(x) + { \tan}^{2} (x)$

$\rm = { \sec}^{2}(x) + { \sec}^{2} (x) - 1$

$\rm = 2{ \sec}^{2} (x) - 1$

= RHS (Hence Proved)

Formula Used:

➡ sin²(x) + cos²(x) = 1

➡ sec²(x) - tan²(x) = 1

Answer 2

Answer:

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