To solve the problem, we need to compute all six trigonometric functions for $\theta$, given that $\csc\theta = \sqrt{6}$ and $\theta$ is an acute angle. Let's proceed step by step:

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Step 1: Understanding $\csc\theta$

$\csc\theta = \frac{1}{\sin\theta}$.
Since $\csc\theta = \sqrt{6}$, we can find $\sin\theta$:
$\sin\theta = \frac{1}{\csc\theta} = \frac{1}{\sqrt{6}}.$
Now, rationalize the denominator: $\sin\theta = \frac{\sqrt{6}}{6}.$

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Step 2: Pythagorean Theorem

For a right triangle, if $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{6}}{6}$, let the opposite side be $\sqrt{6}$ and the hypotenuse be $6$. Then, use the Pythagorean theorem to find the adjacent side:
$\text{Adjacent side} = \sqrt{\text{Hypotenuse}^2 - \text{Opposite}^2}.$
$\text{Adjacent side} = \sqrt{6^2 - (\sqrt{6})^2} = \sqrt{36 - 6} = \sqrt{30}.$
Thus, the adjacent side is $\sqrt{30}$.

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Step 3: Compute Remaining Trigonometric Functions

Part 1: Secant ($\sec\theta$)

$\sec\theta = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{6}{\sqrt{30}}.$ Rationalize the denominator: $\sec\theta = \frac{6\sqrt{30}}{30} = \frac{\sqrt{30}}{5}.$

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Part 2: Cotangent ($\cot\theta$)

$\cot\theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{\sqrt{30}}{\sqrt{6}}.$ Simplify the radicals: $\cot\theta = \sqrt{\frac{30}{6}} = \sqrt{5}.$

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Part 3: Sine ($\sin\theta$)

Already computed: $\sin\theta = \frac{\sqrt{6}}{6}.$

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Part 4: Cosine ($\cos\theta$)

$\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{30}}{6}.$

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Part 5: Tangent ($\tan\theta$)

$\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sqrt{6}}{\sqrt{30}}.$ Simplify the radicals: $\tan\theta = \sqrt{\frac{6}{30}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}.$ Rationalize the denominator: $\tan\theta = \frac{\sqrt{5}}{5}.$

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Final Answers

1. $\sec\theta = \frac{\sqrt{30}}{5}$,
2. $\cot\theta = \sqrt{5}$,
3. $\sin\theta = \frac{\sqrt{6}}{6}$,
4. $\cos\theta = \frac{\sqrt{30}}{6}$,
5. $\tan\theta = \frac{\sqrt{5}}{5}$.

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Would you like me to clarify any step further? 😊

Here are 5 related questions to consider:

1. What is the reciprocal of each trigonometric function?
2. How does $\csc\theta = \sqrt{6}$ relate to the unit circle?
3. How do these values change if $\theta$ is obtuse instead of acute?
4. What are the exact values of these functions if $\csc\theta = 2$?
5. Can we simplify $\tan\theta$ further in terms of sine and cosine?

Tip: Always check if your trigonometric values satisfy the Pythagorean identities!