To solve the equation $\tan^{-1}(\theta) = 2 - \sqrt{3}$ and deduce that $\theta = \tan(\pi/12)$, follow these steps:

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1. Understanding the equation

The given equation is: $\tan^{-1}(\theta) = 2 - \sqrt{3}$ This means that the angle whose tangent is $\theta$ equals $2 - \sqrt{3}$. Therefore: $\theta = \tan(2 - \sqrt{3})$

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2. Relating to a known angle

From trigonometry, you aim to relate $2 - \sqrt{3}$ to a known angle in radians. Observe that the given value $2 - \sqrt{3}$ is not a direct angle but corresponds to a well-known trigonometric relationship at $\pi/12$ radians.

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3. Recognizing the tangent value

From trigonometric tables or properties, we know: $\tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3}$

This is a standard result in trigonometry derived from angle-sum identities: $\tan\left(\frac{\pi}{12}\right) = \tan\left(\frac{\pi}{4} - \frac{\pi}{3}\right)$

Using the tangent subtraction identity: $\tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)}$ Let $a = \frac{\pi}{4}$ and $b = \frac{\pi}{3}$. Then: $\tan\left(\frac{\pi}{12}\right) = \frac{\tan\left(\frac{\pi}{4}\right) - \tan\left(\frac{\pi}{3}\right)}{1 + \tan\left(\frac{\pi}{4}\right)\tan\left(\frac{\pi}{3}\right)}$

Substituting $\tan\left(\frac{\pi}{4}\right) = 1$ and $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$: $\tan\left(\frac{\pi}{12}\right) = \frac{1 - \sqrt{3}}{1 + 1\cdot\sqrt{3}} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}}$

Now simplify: $\tan\left(\frac{\pi}{12}\right) = \frac{(1 - \sqrt{3})(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{1 - 2\sqrt{3} + 3}{1 - 3} = \frac{4 - 2\sqrt{3}}{-2} = 2 - \sqrt{3}$

Thus: $\tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3}$

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4. Conclusion

The equation $\tan^{-1}(\theta) = 2 - \sqrt{3}$ implies: $\theta = \tan(\pi/12)$

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Would you like me to clarify any steps further or explore a different approach?

5 Related Questions:

1. How is $\tan(\pi/12)$ derived from angle subtraction formulas?
2. What are the key steps to simplify tangent expressions involving square roots?
3. How does the identity $\tan(a - b)$ work for $\pi/4 - \pi/3$?
4. Can you show how to verify $\tan^{-1}(2 - \sqrt{3}) = \pi/12$?
5. Why is $\pi/12$ significant in trigonometry?

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Tip: Memorizing common trigonometric values and identities, such as $\tan(\pi/12)$, greatly simplifies solving problems involving inverse functions.