Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

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

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

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

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

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

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

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

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

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

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

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

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

$Now\frac{1}{125}(27{cosec}^6\alpha +8{\sec }^6\alpha )$

$=\frac{1}{125}\left[\right(3{cosec}^2\alpha {)}^3+\left(2{\sec }^2\alpha {)}^3\right]$

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

$=\frac{1}{125}[125+125]$

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

$=2Ans.$