Question

If sin A + sin${}^2$ A + sin${}^3$ A = 1; then find cos${}^6$ A – 4 cos${}^4$ A + 8 cos${}^2$ A -

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

The correct option is A

4

$\begin{array}{c}\sin A+{\sin }^{2}A+{\sin }^{3}A=1\\ \Rightarrow \sin A+{\sin }^{2}A+{\sin }^{3}A={\cos }^{2}A+{\sin }^{2}A.........\left(1={\cos }^{2}A+{\sin }^{2}A\right)\\ \Rightarrow \sin A+{\sin }^{3}A={\cos }^{2}A\\ \Rightarrow \sin A\left(1+{\sin }^{2}A\right)={\cos }^{2}A\\ \Rightarrow \sin A\left(1+1-{\cos }^{2}A\right)={\cos }^{2}A\\ Squaringbothsides-\\ \Rightarrow {\sin }^{2}A{\left(2-{\cos }^{2}A\right)}^2={\cos }^{4}A\\ \Rightarrow \left(1-{\cos }^{2}A\right)\left(4+{\cos }^{4}A-4{\cos }^{2}A\right)={\cos }^{4}A\\ \Rightarrow 4+{\cos }^{4}A-4{\cos }^{2}A-4{\cos }^{2}A-{\cos }^{6}A+4{\cos }^{4}A={\cos }^{4}A\\ \Rightarrow 4={\cos }^{6}A-4{\cos }^{4}A-{\cos }^{4}A+{\cos }^{4}A+4{\cos }^{2}A+4{\cos }^{2}A\\ \Rightarrow {\cos }^{6}A-4{\cos }^{4}A+8{\cos }^{2}A=4\end{array}$