Question

If $15\sin^4\alpha + 10\cos^4\alpha = 6$, for some $\alpha \in R$, then the value of $27\sec^6\alpha + 8\operatorname{cosec}^6\alpha$ is equal to :

• A. 250
• B. 125
• C. 500
• D. 375

Answer 1

Answer: (A)

250

Answer 2

Let $x = \sin^2\alpha$ and $y = \cos^2\alpha$. Given $15\sin^4\alpha + 10\cos^4\alpha = 6$, we have $15x^2 + 10y^2 = 6$. Since $x+y=1$, substitute $y=1-x$: $15x^2 + 10(1-x)^2 = 6$ $15x^2 + 10(1 - 2x + x^2) = 6$ $15x^2 + 10 - 20x + 10x^2 = 6$ $25x^2 - 20x + 4 = 0$ This is a perfect square: $(5x - 2)^2 = 0$, so $x = 2/5$. Thus, $\sin^2\alpha = 2/5$. Then, $\cos^2\alpha = 1 - \sin^2\alpha = 1 - 2/5 = 3/5$.

We need to find $27\sec^6\alpha + 8\operatorname{cosec}^6\alpha$. $\sec^2\alpha = \frac{1}{\cos^2\alpha} = \frac{1}{3/5} = \frac{5}{3}$ $\operatorname{cosec}^2\alpha = \frac{1}{\sin^2\alpha} = \frac{1}{2/5} = \frac{5}{2}$

The expression becomes: $27(\sec^2\alpha)^3 + 8(\operatorname{cosec}^2\alpha)^3$ $= 27\left(\frac{5}{3}\right)^3 + 8\left(\frac{5}{2}\right)^3$ $= 27 \times \frac{125}{27} + 8 \times \frac{125}{8}$ $= 125 + 125 = 250$.