Question : If $\cos A = \sin ^2A$, and $a \sin ^{12}A+b \sin ^{10}A+c \sin ^{8}A+ \sin ^{6}A=1$, then $a + b + c$?
Option 1: 7
Option 2: 8
Option 3: 9
Option 4: 6
Correct Answer: 7
Solution :
$\cos A = \sin^2A$
Square both sides,
$⇒\cos^2A = \sin^4A$
$⇒1 - \sin^2A = \sin^4A$
$⇒1 = \sin^4A + \sin^2A$
Cubing both sides,
$⇒1 =(\sin^4A + \sin^2A)^3$
$⇒1 = \sin^{12}A + \sin^6A + 3\sin^8A + 3\sin^{10}A$
Comparing this with the given expression,
$a \sin^{12}A + b \sin^{10}A + c \sin^8A + \sin^6A = 1$,
We can see that $a = 1$, $b = 3$, and $c = 3$.
So, $a + b + c = 1 + 3 + 3 = 7$
Hence, the correct answer is 7.
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