Question

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

• A. $10$
• B. $8$
• C. $7$
• D. $9$

Answer 1

The correct option is B $8$

$\sin A+{\sin }^{2}A=1$

$\Rightarrow (\sin A=1-{\sin }^{2}A$

$\Rightarrow \sin A={\cos }^{2}A$$(\text{Using}{\cos }^{2}A=1-{\sin }^{2}A)$

$\Rightarrow {\sin }^{2}A={\cos }^{4}A$$\left(\text{on~squaring~both~sides}\right)$

$\Rightarrow 1-{\cos }^{2}A={\cos }^{4}A$

$\Rightarrow 1={\cos }^{2}A+{\cos }^{4}A$

$\Rightarrow 1^3={({\cos }^{2}A+{\cos }^{4}A)}^3$$\left(\text{taking cube of both sides}\right)$

$\Rightarrow 1={\cos }^{12}A+3{\cos }^{10}A+3{\cos }^{8}A+{\cos }^{6}A$

$[\text{Using,}{(a+b)}^3=a^3+3a^{2}b+3a{b}^2+b^3]$

$\Rightarrow 0={\cos }^{12}A+3{\cos }^{10}A+3{\cos }^{8}A+{\cos }^{6}A-1$

$\therefore {\cos }^{12}A+3{\cos }^{10}A+3{\cos }^{8}A+{\cos }^{6}A-1=0$

On comparing the above equation with

$a{\cos }^{12}A+b{\cos }^{10}A+c{\cos }^{8}A+d{\cos }^{6}A-1=0$

we have, $a=1,b=3,c=3,d=1$

$\therefore a+b+c+d=1+3+3+1=8$