Question

Cos⁶A + sin⁶A = 1- K sin²A cos² A where K =​

Answer 1

Answer:

(cos²) ³+(sin²) ³=(sin²+cos²) (sin⁴+cos⁴-sin²cos²)

=1(sin⁴+cos⁴-sin²cos²) =((sin²+cos²) ²-2sin²cos²-sin²cos²)

1²-3sin²cos²

1-3sin²cos²

k=3

Answer 2

$\huge \dag \boxed{ \mathfrak{ \purple{Answer}}}$

$\cos ^{6}a + \sin^{6} a + k \sin {}^{2} 2a = 1$

$\sf\implies \: ( \cos ^{2} a) ^{3} + ( {sin}^{2} a) ^{3} + k \sin ^{2} 2a = 1$

$\sf \implies1 - 3 { \sin }^{2} a \cos ^{2} a + k \: \sin ^{2} 2a = 1$

$\implies \: 1 - \frac{3}{4} { \sin }^{2} 2 a + k \: { \sin }^{2} 2a = 1$

$\implies1 + { \sin }^{2} 2a(k - \frac{3}{4} ) = 1$

$\implies \: k = \frac{3}{4}$