Trigonometry: Solving for $\cot^6\theta + \text{cosec}^6\theta$

We are given the equation: $3(\sin^2\theta + \text{cosec}^2\theta) = 10$

Simplify the Given Equation

1. Divide both sides by 3: $ \sin^2\theta + \text{cosec}^2\theta = \frac{10}{3} $
2. Use the identity $ \sin^2\theta = \frac{1}{\text{cosec}^2\theta} $: $ \frac{1}{\text{cosec}^2\theta} + \text{cosec}^2\theta = \frac{10}{3} $
3. Let $ x = \text{cosec}^2\theta $. The equation becomes: $ \frac{1}{x} + x = \frac{10}{3} $
4. Multiply by $ 3x $ to clear fractions: $ 3 + 3x^2 = 10x $
5. Rearrange into a quadratic equation: $ 3x^2 - 10x + 3 = 0 $
6. Factor the quadratic equation: $ (3x - 1)(x - 3) = 0 $
7. The possible values for $ x $ are $ x = 3 $ or $ x = \frac{1}{3} $.
8. Since $ x = \text{cosec}^2\theta $, and $ \text{cosec}^2\theta \ge 1 $, we must have $ \text{cosec}^2\theta = 3 $.

Calculate $\cot^2\theta$

Use the trigonometric identity $ \cot^2\theta = \text{cosec}^2\theta - 1 $. Substituting the value of $ \text{cosec}^2\theta $: $ \cot^2\theta = 3 - 1 = 2 $

Calculate the Final Expression

We need to find the value of $ \cot^6\theta + \text{cosec}^6\theta $. This can be written as $ (\cot^2\theta)^3 + (\text{cosec}^2\theta)^3 $. Substitute the values we found: $ \cot^2\theta = 2 $ and $ \text{cosec}^2\theta = 3 $. $ (\cot^2\theta)^3 + (\text{cosec}^2\theta)^3 = (2)^3 + (3)^3 $ $ = 8 + 27 $ $ = 35 $

Therefore, the value of $ \cot^6\theta + \text{cosec}^6\theta $ is 35.