Question

$10\sin^4\theta + 15\cos^4\theta = 6$

$\frac{27\operatorname{cosec}^6\theta + 8\sec^6\theta}{16\sec^8\theta}$

Answer 1

$\frac{2}{5}$

Answer 2

To solve the problem, we first need to find the values of $\sin^2\theta$ and $\cos^2\theta$ from the given equation, and then substitute them into the expression to be evaluated.

Step 1: Solve the given equation for $\sin^2\theta$ and $\cos^2\theta$. The given equation is: $10\sin^4\theta + 15\cos^4\theta = 6$

Divide the entire equation by $\cos^4\theta$ (assuming $\cos\theta \neq 0$): $10\frac{\sin^4\theta}{\cos^4\theta} + 15\frac{\cos^4\theta}{\cos^4\theta} = \frac{6}{\cos^4\theta}$ $10\tan^4\theta + 15 = 6\sec^4\theta$

We know the identity $\sec^2\theta = 1 + \tan^2\theta$. So, $\sec^4\theta = (1 + \tan^2\theta)^2$. Substitute this into the equation: $10\tan^4\theta + 15 = 6(1 + \tan^2\theta)^2$

Let $x = \tan^2\theta$. The equation becomes a quadratic in $x$: $10x^2 + 15 = 6(1+x)^2$ $10x^2 + 15 = 6(1 + 2x + x^2)$ $10x^2 + 15 = 6 + 12x + 6x^2$ Rearrange the terms to form a standard quadratic equation: $4x^2 - 12x + 9 = 0$

This quadratic equation is a perfect square: $(2x - 3)^2 = 0$. Solving for $x$: $2x - 3 = 0$ $2x = 3$ $x = \frac{3}{2}$

So, we have $\tan^2\theta = \frac{3}{2}$.

Now, we can find $\sin^2\theta$ and $\cos^2\theta$ using $\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}$ and $\sin^2\theta + \cos^2\theta = 1$. From $\tan^2\theta = \frac{3}{2}$: $\frac{\sin^2\theta}{\cos^2\theta} = \frac{3}{2}$ $2\sin^2\theta = 3\cos^2\theta$

Substitute $\cos^2\theta = 1 - \sin^2\theta$: $2\sin^2\theta = 3(1 - \sin^2\theta)$ $2\sin^2\theta = 3 - 3\sin^2\theta$ $5\sin^2\theta = 3$ $\sin^2\theta = \frac{3}{5}$

Now find $\cos^2\theta$: $\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{3}{5} = \frac{2}{5}$

Step 2: Evaluate the given expression. The expression to evaluate is: $\frac{27\operatorname{cosec}^6\theta + 8\sec^6\theta}{16\sec^8\theta}$

We know that $\operatorname{cosec}\theta = \frac{1}{\sin\theta}$ and $\sec\theta = \frac{1}{\cos\theta}$. So, $\operatorname{cosec}^6\theta = (\sin^2\theta)^{-3}$ and $\sec^6\theta = (\cos^2\theta)^{-3}$ and $\sec^8\theta = (\cos^2\theta)^{-4}$.

Substitute the values of $\sin^2\theta = \frac{3}{5}$ and $\cos^2\theta = \frac{2}{5}$: $= \frac{27\left(\frac{3}{5}\right)^{-3} + 8\left(\frac{2}{5}\right)^{-3}}{16\left(\frac{2}{5}\right)^{-4}}$ $= \frac{27\left(\frac{5}{3}\right)^3 + 8\left(\frac{5}{2}\right)^3}{16\left(\frac{5}{2}\right)^4}$ $= \frac{27 \times \frac{5^3}{3^3} + 8 \times \frac{5^3}{2^3}}{16 \times \frac{5^4}{2^4}}$ $= \frac{27 \times \frac{125}{27} + 8 \times \frac{125}{8}}{16 \times \frac{625}{16}}$ $= \frac{125 + 125}{625}$ $= \frac{250}{625}$

Simplify the fraction: Divide both numerator and denominator by 25: $= \frac{10}{25}$ Divide both numerator and denominator by 5: $= \frac{2}{5}$

The final answer is $\frac{2}{5}$.

The final answer is $\boxed{\frac{2}{5}}$.

Explanation of the solution:

1. Transform the given equation $10\sin^4\theta + 15\cos^4\theta = 6$ into a quadratic equation in terms of $\tan^2\theta$ by dividing by $\cos^4\theta$ and using $\sec^2\theta = 1 + \tan^2\theta$.
2. Solve the quadratic equation $(2\tan^2\theta - 3)^2 = 0$ to find $\tan^2\theta = \frac{3}{2}$.
3. Use $\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}$ and $\sin^2\theta + \cos^2\theta = 1$ to determine $\sin^2\theta = \frac{3}{5}$ and $\cos^2\theta = \frac{2}{5}$.
4. Substitute these values into the expression $\frac{27\operatorname{cosec}^6\theta + 8\sec^6\theta}{16\sec^8\theta}$, rewriting terms as powers of $\sin^2\theta$ and $\cos^2\theta$.
5. Evaluate the numerical expression to get $\frac{250}{625}$, which simplifies to $\frac{2}{5}$.

Answer: $\frac{2}{5}$