Question

The value of $3\frac{si{n}^{4}t+co{s}^{4}t-1}{si{n}^{6}t+co{s}^{6}t-1}$ is equal to k. Find the value of $k$.

Answer 1

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