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. Rewrite the given equation: \[ 10 \sin^4 \theta + 15 \cos^4 \theta = 6 \] \[ 10 (\sin^2 \theta)^2 + 15 (1 - \sin^2 \theta)^2 = 6 \]
2. Let $u = \sin^2 \theta$: \[ 10u^2 + 15(1 - u)^2 = 6 \] \[ 10u^2 + 15(1 - 2u + u^2) = 6 \] \[ 10u^2 + 15 - 30u + 15u^2 = 6 \] \[ 25u^2 - 30u + 9 = 0 \]
3. Solve the quadratic equation: \[ u = \frac{30 \pm \sqrt{900 - 900}}{50} = \frac{30 \pm 0}{50} = \frac{3}{5} \] \[ \sin^2 \theta = \frac{3}{5}, \quad \cos^2 \theta = \frac{2}{5} \]
4. Calculate the given expression: \[ \frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta} = \frac{27 \left( \frac{5}{3} \right)^3 + 8 \left( \frac{5}{2} \right)^3}{16 \left( \frac{5}{2} \right)^4} \] \[ = \frac{27 \cdot \frac{125}{27} + 8 \cdot \frac{125}{8}}{16 \cdot \frac{625}{16}} = \frac{125 + 125}{625} = \frac{250}{625} = \frac{2}{5} \] Therefore, the correct answer is (1) $\frac{2}{5}$.

Answer 2

Given: \(10\sin^4\theta + 15\cos^4\theta = 6\). We must evaluate \[ \frac{27\csc^6\theta + 8\sec^6\theta}{16\sec^8\theta}. \]

Concept Used:

Use \(\sin^2\theta = x\) so that \(\cos^2\theta = 1-x\). Solve the resulting quadratic to find \(x\). Then compute \(\csc^2\theta = 1/x\) and \(\sec^2\theta = 1/(1-x)\) to evaluate the expression.

Step-by-Step Solution:

Step 1: Put \(x=\sin^2\theta\). Then \(\cos^2\theta=1-x\) and the given equation becomes:

\[ 10x^2 + 15(1-x)^2 = 6 \] \[ 10x^2 + 15(1-2x+x^2) = 6 \Rightarrow 25x^2 - 30x + 9 = 0. \]

Step 2: Solve the quadratic:

\[ \Delta = (-30)^2 - 4\cdot 25 \cdot 9 = 900 - 900 = 0 \Rightarrow x=\frac{-(-30)}{2\cdot 25}=\frac{30}{50}=\frac{3}{5}. \] \[ \therefore \sin^2\theta=\frac{3}{5},\quad \cos^2\theta=\frac{2}{5}. \]

Step 3: Compute \(\csc^2\theta\) and \(\sec^2\theta\):

\[ \csc^2\theta=\frac{1}{\sin^2\theta}=\frac{5}{3}\Rightarrow \csc^6\theta=\left(\frac{5}{3}\right)^3=\frac{125}{27}, \] \[ \sec^2\theta=\frac{1}{\cos^2\theta}=\frac{5}{2}\Rightarrow \sec^6\theta=\left(\frac{5}{2}\right)^3=\frac{125}{8},\quad \sec^8\theta=\left(\frac{5}{2}\right)^4=\frac{625}{16}. \]

Step 4: Evaluate the required expression:

\[ \frac{27\csc^6\theta + 8\sec^6\theta}{16\sec^8\theta} =\frac{27\cdot \frac{125}{27} + 8 \cdot \frac{125}{8}}{16\cdot \frac{625}{16}} =\frac{125+125}{625}=\frac{250}{625}=\frac{2}{5}. \]

Final Computation & Result

\(\displaystyle \frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}=\frac{2}{5}\).