If $cos x + cos^{2}x$ = 1, the numerical value of $(sin^{12}x + 3 sin^{10}x + 3 sin^{8}x + sin^{6}x - 1)$ is :
1). -1
2). 2
3). 0
4). 1
cos x + cos2 x = 1
=> cos x = 1 – cos2 x = sin2 x ...(i)
$\therefore$ sin12x + 3 sin10x + 3 sin8x +sin6x – 1
= (sin4 x + sin2 x)3 – 1
= (cos2 x + sin2x)3 – 1 [By (i)]
= 1 – 1 = 0 .
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