[JEE Advanced 1995] $3(\sin x-\cos x)^4+6(\sin x+\cos x)^2+4(\sin^6x+\cos^6x)$ equals
$3(\sin x-\cos x)^4+6(\sin x+\cos x)^2+4(\sin^6x+\cos^6x)$ equals
(a) $11$
(b) $12$
(c) $13$
(d) $14$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
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Correct answer: (c) $13$
With $s=\sin x\cos x$: $3(1-2s)^2+6(1+2s)+4(1-3s^2)=3-12s+12s^2+6+12s+4-12s^2=13.$
JEE Advanced 1995 · Trigonometry — verified solution by the Vidaara Team.
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