Question

If $10{\sin }^4\alpha +15{\cos }^4\alpha =6$, then the value of $81{\mathrm{cosec}}^8\alpha +18{\sec }^8\alpha$ is equal to ___

Answer 1

Answer: 1250

Given: $10{\sin }^4\alpha +15{\cos }^4\alpha =6$
Divide the above equation by 6 on both sides.
$\frac{10{\sin }^4\alpha }{6}+\frac{15{\cos }^4\alpha }{6}=\frac{6}{6}$
$\left(\frac{5}{3}{\sin }^2\alpha \right)\cdot {\sin }^2\alpha +\left(\frac{5}{2}{\cos }^2\alpha \right)\cdot {\cos }^2\alpha =1$
$\because 10{\sin }^4\alpha +15{\cos }^4\alpha =1$
$\frac{5}{3}{\sin }^2\alpha =1\text{ and }\frac{5}{2}{\cos }^2\alpha =1$
$\frac{5}{3}{\mathrm{cosec}}^2\alpha =1\text{ and }\frac{5}{2}{\sec }^2\alpha =1$
Hence, $81{\mathrm{cosec}}^8\alpha +16{\sec }^8\alpha =$ ?
$=81{\left({\mathrm{cosec}}^2\alpha \right)}^4+16{\left({\sec }^{28}\alpha \right)}^4$
$=81{\left(\frac{5}{3}\right)}^4+16{\left(\frac{5}{2}\right)}^4=81\times \frac{625}{81}+16\times \frac{625}{16}$
$=625+625=1250$