Question

questions =
$\cos {}^{4} a - \cos {}^{2} a = { \sin}^{4} a - { \sin }^{2} a$
solve this Plz​

Answer 1

$\huge{\underline{\bf{\purple{Question:-}}}}$

${ \sin }^{4} a - { \cos }^{2} a = { \sin }^{4} a - { \sin}^{2} a$

$\large{\underline{\bf{\red{Answer:-}}}}$

1. we have,

$LHS = { \cos}^{4} A - { \cos }^{2} A$

$\: \: : \implies \sf \: LHS = { \cos}^{2} A( { \cos }^{2} A - 1)$

$: \implies \sf \: LHS = - { \cos }^{2} A(1 - { \cos}^{2} A)$

$: \implies \sf \: LHS = - { \cos }^{2} A \: { \sin}^{2} A = - (1 - { \sin }^{2} A)$

$\sf { \sin }^{2} A = - { \sin}^{2} A + { \sin }^{4} A$

$: \implies \sf \: LHS = { \sin }^{4} A - { \sin }^{2} A = RHS$

$\huge\underline{ \underline{ \mathbb{ { \green{ тн{ \red{คห{ \purple{кร \: }}}}} }}}}$

Answer 2

Answer:

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$\huge\underline{ \mathbb\red{❥A} \green{n} \mathbb\blue{S} \purple{w} \mathbb \orange{E} \pink{r}}\:$

${ \sin }^{4} a - { \cos }^{2} a = { \sin }^{4} a - { \sin}^{2} asin </p><p>4</p><p> a−cos </p><p>2</p><p> a=sin </p><p>4</p><p> a−sin </p><p>2</p><p> a$

we have,

$LHS = { \cos}^{4} A - { \cos }^{2} ALHS=cos </p><p>4</p><p> A−cos </p><p>2</p><p> A$

$\: \: : \implies \sf \: LHS = { \cos}^{2} A( { \cos }^{2} A - 1):⟹LHS=cos </p><p>2</p><p> A(cos </p><p>2</p><p> A−1)$

$: \implies \sf \: LHS = - { \cos }^{2} A(1 - { \cos}^{2} A):⟹LHS=−cos </p><p>2</p><p> A(1−cos </p><p>2</p><p> A)$

$: \implies \sf \: LHS = - { \cos }^{2} A \: { \sin}^{2} A = - (1 - { \sin }^{2} A):⟹LHS=−cos </p><p>2</p><p> Asin </p><p>2</p><p> A=−(1−sin </p><p>2</p><p> A)$

$\sf { \sin }^{2} A = - { \sin}^{2} A + { \sin }^{4} Asin </p><p>2</p><p> A=−sin </p><p>2</p><p> A+sin </p><p>4</p><p> A$

$: \implies \sf \: LHS = { \sin }^{4} A - { \sin }^{2} A = RHS:⟹LHS=sin </p><p>4</p><p> A−sin </p><p>2</p><p> A=RHS$

$\huge{\red{\tt Hope \: it \: helps \: you}}$