Question

$<b>$prove 8th part .

Answer 1

Heya sista

Refer to the attachment


Hope it helps

Answer 2

$\sf{To \ prove \ :}$

$\sf{sin^6 \theta + cos^6 \theta = 1 - 3sin^2 \theta cos^2 \theta}$

$\sf{ L.H.S. -}$

$\sf{= sin^6 \theta + cos^6 \theta}$

$\sf{= (sin^2 \theta)^3 + (cos^2 \theta)^3}$

a³ + b³ = ( a + b )³ - 3ab ( a + b )

Here, a = sin²θ and b = cos²θ

$\sf{= (sin^2 \theta + cos^2 \theta)^3 - 3sin^2 \theta cos^2 \theta (sin^2 \theta +}$
$\sf{cos^2 \theta)}$

sin²θ + cos²θ = 1

$\sf{= (1)^3 - 3sin^2 \theta cos^2 \theta (1)}$

$\sf{= 1 - sin^2 \theta cos^2 \theta}$

$\sf{= R.H.S.}$

$\sf{Hence \ proved}$