Question

If $10{\sin }^4\alpha +15{\cos }^4\alpha =6$, find the value of $27{\mathrm{cosec}}^6\alpha +8{\sec }^6\alpha$.

Answer 1

$10\sin 4\alpha +15\cos 4\alpha =6$

$10\left({\mathrm{Sin}}^4\alpha +{\cos }^4\alpha \right)+5co{s}^4\alpha =6$

$10\left(\right({\sin }^2\alpha +{\cos }^2\alpha )-2{\sin }^2\alpha {\cos }^2\alpha )\pm 6-5{\cos }^4\alpha$

$10(1-2{\sin }^2\alpha {\cos }^2\alpha )=6-5{\cos }^4\alpha$

$4-20{\sin }^2\alpha {\cos }^2\alpha =-5{\cos }^4\alpha$

$4{\sec }^4\alpha -20{\tan }^2\alpha =-5$

$20{\tan }^2\alpha -4{\sec }^4\alpha -5=0$

$20{\tan }^2\alpha -4(1+{\tan }^2\alpha {)}^2-5=0$

$20{\tan }^2\alpha -4(1+{\tan }^4\alpha +2{\tan }^2\alpha )-5=0$

${\tan }^4\alpha -3{\tan }^2\alpha +\frac{9}{4}=0$

$({\tan }^2\alpha -\frac{3}{2}{)}^2=0$

${\tan }^2\alpha =\frac{3}{2}$

$\tan \alpha =\frac{\sqrt{3}}{\sqrt{2}}$

Now $27\cos e{c}^6\alpha +8{\sec }^6\alpha$

$(3\cos e{c}^2\alpha {)}^3+(2{\sec }^2\alpha {)}^3$

$(3\times \frac{5}{3}{)}^3+{\left(2\times \frac{5}{2}\right)}^3$

$125+125$

$=250$