Question

The expression $\frac{{\sin }^{4}t+{\cos }^{4}t-1}{{\sin }^{6}t+{\cos }^{6}t-1}$ when simplified reduces to .......................

• A. $\frac{2}{3}$
• B. 1
• C. $\frac{1}{3}$
• D. $\frac{2}{5}$

Answer 1

Answer: (A)

$\frac{2}{3}$

$\frac{{\sin }^{4}t+{\cos }^{4}t-1}{{\sin }^{6}t+{\cos }^{6}t-1}=\frac{{({\sin }^{2}t+{\cos }^{2}t)}^2-2{\sin }^{2}t{\cos }^{2}t-1}{{({\sin }^{2}t+{\cos }^{2}t)}^3-3{\sin }^{2}t{\cos }^{2}t({\sin }^{2}t+{\cos }^{2}t)-1}$

since $(a+b{)}^2=a^2+2ab+b^2$ and $(a+b{)}^3=a^3+3a^{2}b+3ab(a+b)$

$=\frac{-2{\sin }^{2}t{\cos }^{2}t}{-3{\sin }^{2}t{\cos }^{2}t}=\frac{2}{3}$