Question

If $10{\sin }^{4}x+15{\cos }^{4}x=6$, then the numerical value of $24{\sec }^{6}x-27{\text{cosec}}^{6}x$ is

Answer 1

$10{\sin }^{4}x+15{\cos }^{4}x=6\Rightarrow 10{\sin }^{4}x+15{\cos }^{4}x=6\times (1{)}^2\Rightarrow 10{\sin }^{4}x+15{\cos }^{4}x=6({\sin }^{2}x+{\cos }^{2}x{)}^2$
Dividing both sides by ${\cos }^{4}x$ we get
$\Rightarrow 10{\tan }^{4}x+15=6({\tan }^{2}x+1{)}^2\Rightarrow 4{\tan }^{4}x-12{\tan }^{2}x+9=0\Rightarrow (2{\tan }^{2}x-3{)}^2=0\Rightarrow {\tan }^{2}x=\frac{3}{2}\Rightarrow {\cot }^{2}x=\frac{2}{3}$
Now,
$24{\sec }^{6}x-27{\text{cosec}}^{6}x=24({\sec }^{2}x{)}^3-27({\text{cosec}}^{2}x{)}^3=24(1+{\tan }^{2}x{)}^3-27(1+{\cot }^{2}x{)}^3=24{\left(\frac{5}{2}\right)}^3-27{\left(\frac{5}{3}\right)}^3=5^3(3-1)=250$