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Express as a function of a DIFFERENT angle, 0^(@) <= theta < 360^(@). tan(310^(@)) tan(◻^(@))

Express as a function of a DIFFERENT angle, 0∘≤θ<360∘ 0^{\circ} \leq \theta<360^{\circ} .\newlinetan⁡(310∘) \tan \left(310^{\circ}\right) \newlinetan⁡(□∘) \tan \left(\square^{\circ}\right)

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Q. Express as a function of a DIFFERENT angle, 0∘≤θ<360∘ 0^{\circ} \leq \theta<360^{\circ} .\newlinetan⁡(310∘) \tan \left(310^{\circ}\right) \newlinetan⁡(□∘) \tan \left(\square^{\circ}\right)
  1. Understand the problem: Understand the problem.\newlineWe need to express tan⁡(310°)\tan(310°) as a function of a different angle within the range of 0°0° to 360°360°. We can use the fact that the tangent function has a period of 180°180°, which means tan⁡(θ)=tan⁡(θ+180°)\tan(\theta) = \tan(\theta + 180°). We can also use the symmetry properties of the tangent function on the unit circle.
  2. Find the reference angle: Find the reference angle.\newlineThe reference angle is the acute angle that the terminal side of the given angle makes with the x-axis. For an angle in the fourth quadrant, like 310∘310^\circ, the reference angle is 360∘−310∘=50∘360^\circ - 310^\circ = 50^\circ. This means tan⁡(310∘)\tan(310^\circ) is equal to tan⁡(50∘)\tan(50^\circ), but since 310∘310^\circ is in the fourth quadrant where tangent is negative, we have tan⁡(310∘)=−tan⁡(50∘)\tan(310^\circ) = -\tan(50^\circ).
  3. Express as a function: Express tan⁡(310°)\tan(310°) as a function of a different angle.\newlineWe can use the reference angle found in Step 22 to express tan⁡(310°)\tan(310°) as a function of 50°50°. Since tan⁡(θ)=−tan⁡(180°−θ)\tan(\theta) = -\tan(180° - \theta) for angles in the second quadrant, we can say tan⁡(310°)=−tan⁡(180°−50°)=−tan⁡(130°)\tan(310°) = -\tan(180° - 50°) = -\tan(130°). Therefore, tan⁡(310°)\tan(310°) can be expressed as −tan⁡(130°)-\tan(130°).

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