Question

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

Answer 1

Given that, $10{\sin }^4\alpha +15{\cos }^4\alpha =6$ ……. (1)

We know that,

${\cos }^2\alpha =1-{\sin }^2\alpha$

Put in equation (1) and we get,

$10{\sin }^4\alpha +15{(1-{\sin }^2\alpha )}^2=6$

$\Rightarrow 10{\sin }^4\alpha +15(1+{\sin }^4\alpha -2{\sin }^2\alpha )=6$

$\Rightarrow 10{\sin }^4\alpha +15+15{\sin }^4\alpha -30{\sin }^2\alpha =6$

$\Rightarrow 25{\sin }^4\alpha -30{\sin }^2\alpha =6-15$

$\Rightarrow 25{\sin }^4\alpha -30{\sin }^2\alpha +9=0$

Let $x=\sin \alpha$ become a quadratic equation.

We get,

$\Rightarrow 25x^4-30x^2+9=0$

$\Rightarrow {(5x^2-3)}^2=0$

$\Rightarrow 5x^2-3=0$

$\Rightarrow x^2=\frac{3}{5}$

$\Rightarrow x=\pm \sqrt{\frac{3}{5}}$

Now, $x=\sin \alpha =\pm \sqrt{\frac{3}{5}}$

So, $\cos ec\alpha =\pm \sqrt{\frac{5}{3}}$

Then, $\sec \alpha =\pm \sqrt{\frac{5}{2}}$ (By Pythagoras theorem)

Now,

Find the value of

$\frac{1}{125}(27\cos e{c}^6\alpha +8{\sec }^6\alpha )$

$=\frac{1}{125}(27\times {\left(\sqrt{\frac{5}{3}}\right)}^6+8\times {\left(\sqrt{\frac{5}{2}}\right)}^6)$

$=\frac{1}{125}(27\times {\left(\frac{5}{3}\right)}^3+8\times {\left(\frac{5}{2}\right)}^3)$

$=\frac{1}{125}(27\times \frac{125}{27}+8\times \frac{125}{8})$

$=\frac{1}{125}(125+125)$

$=\frac{250}{125}$

$=2$

Hence, this is the answer.