Question

$\sf{ {sin}^{4} \: \theta \: - {cos}^{4} \: \theta = 1 - 2 \: {cos}^{2} \: \theta }$

Prove this......
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Answer 1

Answer:

$Required Answer:-</p><p></p><p>Given to prove:</p><p></p><p>( sec θ - cos θ ) ( cot θ + tan θ ) = tan θ sec θ</p><p></p><p>Proof:</p><p></p><p>Taking LHS,</p><p></p><p>\rm ( \sec\theta - \cos \theta)( \cot \theta + \tan \theta)(secθ−cosθ)(cotθ+tanθ)</p><p></p><p>\rm = \bigg( \dfrac{1}{ \cos \theta} - \cos \theta \bigg) \bigg( \dfrac{ \cos \theta }{ \sin \theta } + \dfrac{ \sin \theta}{ \cos \theta} \bigg)=(cosθ1−cosθ)(sinθcosθ+cosθsinθ)</p><p></p><p>\rm = \bigg( \dfrac{1 - \cos ^{2} \theta }{ \cos \theta} \bigg) \bigg( \dfrac{ \cos^{2} \theta + { \sin }^{2} \theta}{ \sin \theta \cos \theta } \bigg)=(cosθ1−cos2θ)(sinθcosθcos2θ+sin2θ)</p><p></p><p>\rm = \dfrac{ \sin^{2} \theta }{ \cos \theta} \times \dfrac{1}{ \sin \theta \cos \theta }=cosθsin2θ×sinθcosθ1</p><p></p><p>\rm = \dfrac{ \sin \theta }{ \cos \theta} \times \dfrac{1}{ \cos \theta }=cosθsinθ×cosθ1</p><p></p><p>\rm = \tan \theta\times \dfrac{1}{ \cos \theta }=tanθ×cosθ1</p><p></p><p>\rm = \tan \theta \sec \theta=tanθsecθ</p><p></p><p>= RHS (Hence Proved)</p><p></p><p>Formula Used:</p><p></p><p>sin(x)/cos(x) = tan(x)</p><p></p><p>cos(x)/sin(x) = cot(x)</p><p></p><p>sin²(x) + cos²(x) = 1</p><p></p><p>cos(x) = 1/sec(x)</p><p></p><p>sec(x) = 1/cos(x)</p><p></p><p>$

Answer 2

Answer:

Answer:

Rt Hon Boris Johnson MP is the answer