<s>[INST] If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$ If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$.
Find the value of $2b + \dfrac {c}{a}$.
My Attempt:
$$\sin A+\sin^2 A=1$$
$$\sin A + 1 - \cos^2 A=1$$
$$\sin A=\cos^2 A$$
Now,
$$a(\cos^2 A)^{6}+b(\cos^2 A)^{4}+c(\cos^2 A)^{3}=1$$
$$a\sin^6 A+ b\sin^4 A+c\sin^3 A=1$$
How do I proceed further?
 [/INST] @MyGlasses .. In the third line from below you forgot that there is $(-a)$ twice in your attempt to prove the required identity, resulting in a mistake in your answer. 
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