Question

please tell me answer​

Answer 1

sinA + cosA = a

=> (sinA + cosA)² = a²

=> a² = sin²A + cos²A + 2sinAcosA

=> a² = 1 + 2sinAcosA ... (i) ... (sin²A + cos²A = 1)

now,

R.H.S.

$= 1 - \frac{3}{4}{({a}^{2} - 1)}^{2} \\</p><p>= 1 - \frac{3}{4}{(1 + 2sinA.cosA - 1)}^{2} \\</p><p>= 1 - \frac{3}{4}{(2sinA.cosA)}^{2} \\</p><p>= 1 - \frac{3}{4}(4sin{}^{2} A.cos{}^{2}) \\</p><p>= 1 - 3sin{}^{2}A.cos{}^{2}A$

L.H.S.

$= sin{}^{6}A + cos{}^{6}A \\</p><p>= {(sin{}^{2}A)}^{3} + {(cos{}^{2}A)}^{3}$

= (sin²A + cos²A)³ - 3sin²Acos²A(sin²A + cos²A)

... a³ + b³ = (a + b)³ - 3ab(a + b)

= (1)³ - 3sin²Acos²A(1)

= 1 - 3sin²Acos²A

= R.H.S.

L.H.S. = R.H.S.

Hence,

$sin{}^{6}A + cos{}^{6}A = 1 - \frac{3}{4}{(a{}^{2} - 1)}^{2}$

... Hence Proved!