To solve the expression

$\sec^2\frac{14\pi}{17} - \sec^2\frac{39\pi}{17} + \cot^2\frac{41\pi}{34} - \cot^2\frac{23\pi}{34},$

we will use trigonometric identities and periodicity.

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Step 1: Simplify using periodicity

• Periodicity of $\sec^2 x$: $\sec^2 x$ has a period of $\pi$. Therefore: $\sec^2\frac{39\pi}{17} = \sec^2\left(\frac{39\pi}{17} - 2\pi\right) = \sec^2\frac{5\pi}{17}.$

• Periodicity of $\cot^2 x$: $\cot^2 x$ has a period of $\pi$. Therefore: $\cot^2\frac{41\pi}{34} = \cot^2\left(\frac{41\pi}{34} - \pi\right) = \cot^2\frac{7\pi}{34}.$

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Step 2: Use trigonometric identities

• From the Pythagorean identity: $\sec^2 x = 1 + \tan^2 x, \quad \cot^2 x = \csc^2 x - 1.$

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Step 3: Rewrite the expression

Using the periodicity and identities, the expression becomes: $\sec^2\frac{14\pi}{17} - \sec^2\frac{5\pi}{17} + \cot^2\frac{7\pi}{34} - \cot^2\frac{23\pi}{34}.$

Substitute $\sec^2 x = 1 + \tan^2 x$ and $\cot^2 x = \csc^2 x - 1$: $(1 + \tan^2\frac{14\pi}{17}) - (1 + \tan^2\frac{5\pi}{17}) + (\csc^2\frac{7\pi}{34} - 1) - (\csc^2\frac{23\pi}{34} - 1).$

Simplify: $\tan^2\frac{14\pi}{17} - \tan^2\frac{5\pi}{17} + \csc^2\frac{7\pi}{34} - \csc^2\frac{23\pi}{34}.$

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Step 4: Use symmetry of trigonometric functions

• $\tan^2 x$ and $\csc^2 x$ are symmetric in certain intervals. Further simplification depends on precise angle relationships.

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Related Questions:

1. What are the periodic properties of $\sec x$ and $\cot x$?
2. How can trigonometric identities help simplify similar expressions?
3. What are the differences in the periodicity of $\tan x$ and $\csc x$?
4. How does angle reduction help simplify trigonometric functions?
5. What are other methods to handle higher angles in trigonometric problems?

Tip: Always reduce angles using periodic properties before applying trigonometric identities for easier simplification.