Question

show that
sin^6a + cos^6a + 3 sin²a cos² a = 1 ?​

Answer 1

Answer:

$\huge\bf\underline\purple{AnSweR:}$

${ \sin }^{6} \alpha + { \cos }^{6} \alpha + 3 { \sin}^{2} \alpha { \cos}^{2} \alpha$

$= ( { { \sin}^{2} \alpha + { \cos}^{2} \alpha })^{3} - 3 { \sin }^{2} \alpha { \cos}^{2} \alpha ( { \sin}^{2} \alpha + { \cos }^{2} \alpha ) + 3 \: { \sin }^{2} \alpha \: { \cos}^{2} \alpha$

$= 1 - 3 \: { \sin}^{2} \alpha \: { \cos }^{2} \alpha \: + 3 \: { \sin}^{2} \alpha \: { \cos}^{2} \alpha$

$\implies\bf\red{1}$

Answer 2

Answer:

heya mate

sin^6A+cos^6A

(sin²A)³+(cos²A)³ ............[a^6=(a²)³]

(sin²A+cos²A)³-3sin²Acos²A.........[(a+b)³=a³+b³+3ab]

1³-3sin²Acos²A

1-3sin²Acos²A

!!hence proved!!