Question

If sinA + sin^2A = 1 then cos^12A + 3cos^10A + 3cos^8A +cos^6A = ?​

Answer 1

Given,

$\longrightarrow\sin A+\sin^2A=1$

$\longrightarrow\sin A=1-\sin^2A$

$\longrightarrow\sin A=\cos^2A\quad\quad\dots(1)$

So,

$\longrightarrow S=\cos^{12}A+3\cos^{10}A+3\cos^8A+\cos^6A$

Taking $\cos^6A$ common in RHS,

$\longrightarrow S=\cos^6A\left(\cos^6A+3\cos^4A+3\cos^2A+1\right)$

Factorising RHS,

$\longrightarrow S=\cos^6A\left((\cos^2A)^3+3(\cos^2A)^2\cdot1+3\cos^2A\cdot1^2+1^3\right)$

$\longrightarrow S=(\cos^2A)^3\left(\cos^2A+1\right)^3$

$\longrightarrow S=\left[\cos^2A\left(\cos^2A+1\right)\right]^3$

$\longrightarrow S=\left(\cos^4A+\cos^2A\right)^3$

$\longrightarrow S=\left((\cos^2A)^2+\cos^2A\right)^3$

From (1),

$\longrightarrow S=\left(\sin^2A+\cos^2A\right)^3$

$\longrightarrow\underline{\underline{S=1}}$

Answer 2

Answer:

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