Question

Find the value of
$\sin ^{4} \alpha - \cos ^{4} \alpha =$


Please help no scam ... no copy-paste from Google ❤❤​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

Identity used here :-

$= {x}^{2} - {y}^{2} = (x + y)(x - y)$

$= >{sin}^{4} \alpha - {cos}^{4} \alpha$

We can write $4 \: as \: {(2)}^{2}$

$= > { {(sin}^{2} \alpha ) }^{2} - { {(cos \alpha )}^{2} }^{2}$

$\bold{= ( {sin}^{2} \alpha + {cos}^{2} \alpha )( {sin}^{2} \alpha - {cos}^{2} \alpha )}$

$∴ {sin}^{4} \alpha - {cos}^{4} \alpha$$\bold{= ( {sin}^{2} \alpha + {cos}^{2} \alpha )( {sin}^{2} \alpha - {cos}^{2} \alpha )}$

Other Identity related to this :

$\bold{\boxed{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\bold{\boxed{ {x}^{3} + {y}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

$\bold{\boxed{(x + y)(x + z) = {x}^{2} + (y+ z)x + yz}}$

Answer 2

Answer:

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