If cos x+cos^2 x=1 then the value of sin^12 x+3 sin^10 x+3 sin^8 x+sin^6 x-1 is equal to

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MARKS Content Team · Accepted Answer

Correct answer: (D) 0. Given : cos x+cos^2 x=1 ⇒ cos x=1-cos^2 x ⇒ cos x=sin^2 x ⇒ cos^2 x=sin^4 x ⇒ 1-sin^2 x=sin^4 x ⇒ sin^4 x+sin^2 x=1 cubic both sides, we have sin^12 x+sin^6 x+3 sin^6 x (sin^4 x+sin^2 x )=1 ⇒ sin^12 x+sin^6 x+3 sin^10 x+3 sin^8 x=1 ⇒ sin^12 x+3 sin^10 x+3 sin^8 x+sin^6 x-1=0

If $\cos x+\cos ^{2} x=1$ then the value of $\sin ^{12} x+3 \sin ^{10} x+3 \sin ^{8} x+\sin ^{6} x-1$ is equal to

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