Question

$\text{If}\sin A+{\sin }^{2}A=1\&a{\cos }^{12}A+b{\cos }^{10}A+c{\cos }^{8}A+d{\cos }^{6}A-1=0\text{then}a+b+c+d=$

• A. $10$
• B. $8$
• C. $7$
• D. $9$

Answer 1

The correct option is B $8$

$\sin A+{\sin }^{2}A=1\Rightarrow \sin A={\cos }^{2}A\Rightarrow {\sin }^{2}A={\cos }^{4}A\Rightarrow 1={\cos }^{2}A+{\cos }^{4}A\Rightarrow 1^3={({\cos }^{2}A+{\cos }^{4}A)}^3\Rightarrow 1={\cos }^{12}A+3{\cos }^{10}A+3{\cos }^{8}A+{\cos }^{6}A\Rightarrow 0={\cos }^{12}A+3{\cos }^{10}A+3{\cos }^{8}A+{\cos }^{6}A-1$
From here $a=1,b=3,c=3,d=1\therefore a+b+c+d=8$