Question

$\frac{{\sin }^{4}x+{\cos }^{4}x-1}{{\sin }^{6}x+{\cos }^{6}x-1}$ is simplified to

• A. $0$
• B. $1$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$

Answer 1

The correct option is C $\frac{2}{3}$

$\frac{si{n}^{4}x+co{s}^{4}x-1}{si{n}^{6}x+co{s}^{6}x-1}=\frac{(si{n}^{2}x+co{s}^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x-1}{(si{n}^{2}x+co{s}^{2}x)({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x)-1}=\frac{(1{)}^2-2{\sin }^{2}x{\cos }^{2}x-1}{\left(1\right)\left(\right({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x)-1}=\frac{1-2{\sin }^{2}x{\cos }^{2}x-1}{1-3{\sin }^{2}x{\cos }^{2}x-1}=\frac{-2{\sin }^{2}x{\cos }^{2}x}{-3{\sin }^{2}x{\cos }^{2}x}=\frac{2}{3}$

Therefore, Answer is $\frac{2}{3}$