Question

If $10 \sin^4 \theta + 15 \cos^4 \theta = 6$, then the value of $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is:

• A. $\frac{2}{5}$
• B. $\frac{3}{4}$
• C. $\frac{3}{5}$
• D. $\frac{1}{5}$

Hint:

Rewrite trigonometric expressions in terms of $\sin^2 \theta$ and $\cos^2 \theta$ to simplify calculations.

Answer 1

The Correct Option is A

1. Re-express the given equation: \[ 10 \sin^4 \theta + 15 \cos^4 \theta = 6 \] Substituting $\cos^2 \theta = 1 - \sin^2 \theta$: \[ 10 (\sin^2 \theta)^2 + 15 (1 - \sin^2 \theta)^2 = 6 \] 2. Let $u = \sin^2 \theta$: \[ 10u^2 + 15(1 - u)^2 = 6 \] Expand and simplify: \[ 10u^2 + 15(1 - 2u + u^2) = 6 \] \[ 10u^2 + 15 - 30u + 15u^2 = 6 \] Combine terms to form a quadratic equation: \[ 25u^2 - 30u + 9 = 0 \] 3. Solve the quadratic equation for $u$: \[ u = \frac{-(-30) \pm \sqrt{(-30)^2 - 4(25)(9)}}{2(25)} = \frac{30 \pm \sqrt{900 - 900}}{50} = \frac{30 \pm 0}{50} = \frac{3}{5} \] Thus, we have: \[ \sin^2 \theta = \frac{3}{5}, \quad \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{3}{5} = \frac{2}{5} \] 4. Evaluate the given expression using the derived values of $\sin^2 \theta$ and $\cos^2 \theta$: Recall that $\csc \theta = \frac{1}{\sin \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$. Therefore, $\csc^2 \theta = \frac{1}{\sin^2 \theta} = \frac{5}{3}$ and $\sec^2 \theta = \frac{1}{\cos^2 \theta} = \frac{5}{2}$. \[ \frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta} = \frac{27 (\csc^2 \theta)^3 + 8 (\sec^2 \theta)^3}{16 (\sec^2 \theta)^4} = \frac{27 \left( \frac{5}{3} \right)^3 + 8 \left( \frac{5}{2} \right)^3}{16 \left( \frac{5}{2} \right)^4} \] Substitute the values and simplify: \[ = \frac{27 \cdot \frac{125}{27} + 8 \cdot \frac{125}{8}}{16 \cdot \frac{625}{16}} = \frac{125 + 125}{625} = \frac{250}{625} = \frac{2}{5} \] The final result is $\frac{2}{5}$.