Question

If $10{\sin }^4\alpha +15{\cos }^4\alpha =6$ then find the value of $\frac{1}{125}(27\cos e{c}^6\alpha +8{\sec }^6\alpha )$.

Answer 1

$10{\sin }^4\alpha +15{\cos }^4\alpha =6$

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

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

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

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

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

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

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

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

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

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

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

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

$\text{Now,}\frac{1}{125}\left(27\cos e{c}^6\alpha +8{\sec }^6\alpha \right)\mathrm{Cos}es\alpha =\frac{\sqrt{5}}{\sqrt{3}}=\sqrt{\frac{5}{3}}$

$=\frac{1}{125}\left[{\left(3\cos e{c}^2\alpha \right)}^3+{\left(2{\sec }^2\alpha \right)}^3\right]\sec \alpha =\sqrt{\left(\frac{5}{2}\right)}$

$=\frac{1}{125}\left[{\left(3\times \frac{5}{3}\right)}^3+{\left(2\times \frac{5}{2}\right)}^3\right]$

$\frac{1}{125}\left(125+125\right)$

$\frac{1}{125}\left(250\right)$

$2Ans.$