<Question>
Find the degree measure of the least positive angle $\theta$ for which
\[\tan \theta = \frac{\cos 5^\circ \cos 20^\circ + \cos 35^\circ \cos 50^\circ - \sin 5^\circ \sin 20^\circ - \sin 35^\circ \sin 50^\circ}{\sin 5^\circ \cos 20^\circ - \sin 35^\circ \cos 50^\circ + \cos 5^\circ \sin 20^\circ - \cos 35^\circ \sin 50^\circ}.\]
</Question>

<think>
<Plan_1>
Okay, so I need to find the least positive angle θ in degrees where the tangent of θ is equal to this big expression. The expression is a fraction with a numerator and a denominator, both of which are combinations of cosines and sines of different angles. Hmm, this looks a bit complicated, but maybe I can simplify it using some trigonometric identities. Let me recall the sum and difference formulas, product-to-sum formulas, and see which ones might apply here.

First, let me write down the given expression to have a clear view:

tanθ = [cos5° cos20° + cos35° cos50° - sin5° sin20° - sin35° sin50°] / [sin5° cos20° - sin35° cos50° + cos5° sin20° - cos35° sin50°]

Looking at the numerator: cos5° cos20° + cos35° cos50° - sin5° sin20° - sin35° sin50°

<prompt_1.1> Analyze the structure of the numerator and denominator and identify potential simplifications using trigonometric identities. </prompt_1.1>

Hmm, that seems manageable. Let me check that step again.

<prompt_1.2> Verify the simplification of the numerator using the above plan. </prompt_1.2>

Now moving on to the denominator: sin5° cos20° - sin35° cos50° + cos5° sin20° - cos35° sin50°

Let's rearrange the terms to group similar ones:

(sin5° cos20° + cos5° sin20°) - (sin35° cos50° + cos35° sin50°)

Again, this looks like the sine addition formula. The identity is sin(A + B) = sinA cosB + cosA sinB. So each group is a sine of a sum.

<prompt_1.3> Simplify the denominator using the above plan. </prompt_1.3>
-----

But wait, let me check if there's a possibility of another angle with a smaller positive measure. Since tangent has a period of 180°, but we need the least positive angle. 25° is the answer here. Let me confirm once again.

<prompt_1.4> Confirm the final result and ensure it is the least positive angle. </prompt_1.4>
</Plan_1>

<Execution_1>
<execution_1.1>
I notice that the terms can be grouped into two products each. The first two terms are cos5° cos20° and cos35° cos50°, and then the last two terms are -sin5° sin20° and -sin35° sin50°. Maybe I can apply the product-to-sum identities here. Recall that cosA cosB = [cos(A+B) + cos(A-B)] / 2 and sinA sinB = [cos(A-B) - cos(A+B)] / 2. Let me try applying these identities to each pair.

Starting with the first term, cos5° cos20°:

cos5° cos20° = [cos(25°) + cos(-15°)] / 2 = [cos25° + cos15°] / 2, since cosine is even.

Similarly, cos35° cos50°:

cos35° cos50° = [cos85° + cos(-15°)] / 2 = [cos85° + cos15°] / 2.

Now, the sine terms: -sin5° sin20°. Let's apply the identity sinA sinB = [cos(A-B) - cos(A+B)] / 2.

-sin5° sin20° = -[cos(-15°) - cos25°] / 2 = -[cos15° - cos25°] / 2 = [cos25° - cos15°] / 2.

Similarly, -sin35° sin50°:

-sin35° sin50° = -[cos(-15°) - cos85°] / 2 = -[cos15° - cos85°] / 2 = [cos85° - cos15°] / 2.

So putting all the numerator terms together:

[cos25° + cos15°]/2 + [cos85° + cos15°]/2 + [cos25° - cos15°]/2 + [cos85° - cos15°]/2

Let me combine these fractions:

= [cos25° + cos15° + cos85° + cos15° + cos25° - cos15° + cos85° - cos15°] / 2

Simplify the numerator:

cos25° + cos25° + cos85° + cos85° + cos15° + cos15° - cos15° - cos15°

Which simplifies to:

2cos25° + 2cos85°
</execution_1.1>

<execution_1.2>
The numerator has two products of cosines and two products of sines, each with a minus sign. The first two terms are cos5° cos20°, and the last two are -sin5° sin20°, but with different angles. Wait, cos5° cos20° - sin5° sin20° is equal to cos(5° + 20°) = cos25°, right? Similarly, cos35° cos50° - sin35° sin50° is cos(35° + 50°) = cos85°. Let me verify that:

Using the cosine addition formula: cos(A + B) = cosA cosB - sinA sinB. So yes, if we have cos5° cos20° - sin5° sin20°, that's cos(5° + 20°) = cos25°. Similarly, cos35° cos50° - sin35° sin50° = cos(35° + 50°) = cos85°. So the numerator simplifies to cos25° + cos85°.
</execution_1.2>

<execution_1.3>
So:

sin5° cos20° + cos5° sin20° = sin(5° + 20°) = sin25°

Similarly, sin35° cos50° + cos35° sin50° = sin(35° + 50°) = sin85°

Therefore, the denominator simplifies to sin25° - sin85°
</execution_1.3>

<execution_1.4>
The simplification steps led us to θ = 25°, and since 25° is already between 0° and 180°, and there's no smaller positive angle that would satisfy tanθ equal to that expression. Therefore, 25° is indeed the least positive solution.
</execution_1.4>
</Execution_1>

<Plan_2>
Based on execution_1.3:
So now, our original expression simplifies to:

tanθ = [cos25° + cos85°] / [sin25° - sin85°]

Now, I need to simplify this further. Let's see if we can combine the terms in the numerator and the denominator. Maybe using sum-to-product identities again.

<prompt_2.1> Simplify the expression further using sum-to-product identities as per the above plan. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Starting with the numerator: cos25° + cos85°. The sum-to-product identity for cosines is cosA + cosB = 2cos[(A+B)/2] cos[(A-B)/2].

Applying that:

cos25° + cos85° = 2cos[(25+85)/2] cos[(85-25)/2] = 2cos55° cos30°

Similarly, the denominator: sin25° - sin85°. The sum-to-product identity for sine subtraction is sinA - sinB = 2cos[(A+B)/2] sin[(A-B)/2]. So:

sin25° - sin85° = 2cos[(25+85)/2] sin[(25-85)/2] = 2cos55° sin(-30°) = -2cos55° sin30°

Since sin(-θ) = -sinθ.

So putting numerator and denominator together:

tanθ = [2cos55° cos30°] / [-2cos55° sin30°]

The 2cos55° terms cancel out from numerator and denominator, leaving:

tanθ = [cos30°] / [-sin30°] = -cos30° / sin30°

But cos30° is √3/2 and sin30° is 1/2, so:

tanθ = -(√3/2) / (1/2) = -√3

Therefore, tanθ = -√3. But we need the least positive angle θ where this holds. However, tangent is negative in the second and fourth quadrants. Since we want the least positive angle, it must be in the second quadrant because the fourth quadrant angle would be larger than 360° - θ, but let's check.

We know that tanθ = -√3. The reference angle where tan is √3 is 60°, so the angles where tanθ = -√3 are 180° - 60° = 120° and 360° - 60° = 300°. But since we need the least positive angle, 120° is smaller than 300°, so θ = 120°.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, but let me double-check my steps because 120° seems a bit large. Let me verify each step again.

<prompt_3.1> Verify the simplification steps to ensure there are no errors as per the above plan. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Starting from the original expression, after simplifying numerator and denominator:

Numerator: cos25 + cos85 = 2cos55 cos30 (correct)

Denominator: sin25 - sin85 = -2cos55 sin30 (correct)

Thus, tanθ = (2cos55 cos30) / (-2cos55 sin30) = -cos30 / sin30 = -√3 (since cos30 = √3/2, sin30 = 1/2, so (√3/2)/(1/2) = √3, but with a negative sign, so -√3). Therefore, tanθ = -√3.

The solutions to tanθ = -√3 are θ = 120° + k*180°, where k is integer. The least positive angle here is 120°, since 120° - 180° = -60°, which is negative, so 120° is the smallest positive.
</execution_3.1>
</Execution_3>

<Final_answer>
Therefore, the least positive angle $\theta$ for which $\tan \theta = -\sqrt{3}$ is $\boxed{120^\circ}$.
</Final_answer>
</think>