Question

If 15 sin⁴ α + 10 cos⁴ α = 6, for some α ∈ R, then the value of 27 sec⁶ α + 8 cosec⁶ α is equal to :

• A. 500
• B. 400
• C. 350
• D. 250

Hint:

Converting trigonometric equations into a single function (like $\tan \alpha$) often reveals a perfect square quadratic.

Answer 1

The Correct Option is D

Step 1: Given $15 \sin^4 \alpha + 10 \cos^4 \alpha = 6$. Divide by $\cos^4 \alpha$: $15 \tan^4 \alpha + 10 = 6 \sec^4 \alpha = 6(1 + \tan^2 \alpha)^2$. $15 \tan^4 \alpha + 10 = 6(1 + 2\tan^2 \alpha + \tan^4 \alpha)$. $9 \tan^4 \alpha - 12 \tan^2 \alpha + 4 = 0 \implies (3 \tan^2 \alpha - 2)^2 = 0$. So, $\tan^2 \alpha = \frac{2}{3}$.
Step 2: Then $\sin^2 \alpha = \frac{2}{5}$ and $\cos^2 \alpha = \frac{3}{5}$. Therefore, $\csc^2 \alpha = \frac{5}{2}$ and $\sec^2 \alpha = \frac{5}{3}$.
Step 3: Value $= 27(\sec^2 \alpha)^3 + 8(\csc^2 \alpha)^3 = 27(\frac{5}{3})^3 + 8(\frac{5}{2})^3 = 27(\frac{125}{27}) + 8(\frac{125}{8}) = 125 + 125 = 250$.