<Question>
Simplify $\tan 100^\circ + 4 \sin 100^\circ.$
</Question>

<think>
<Plan_1>
Okay, so I need to simplify the expression tan 100° + 4 sin 100°. Hmm, let me start by recalling some trigonometric identities and see if I can manipulate this expression into a more straightforward form.

First, I know that tan θ is equal to sin θ over cos θ. So maybe I can rewrite tan 100° as sin 100° divided by cos 100°. Let me try that:

<prompt_1.1> Rewrite the expression tan 100° + 4 sin 100° using the identity tan θ = sin θ / cos θ as per the above plan. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
tan 100° + 4 sin 100° = (sin 100° / cos 100°) + 4 sin 100°.

Hmm, now both terms have sin 100°, so maybe I can factor that out? Let's see:

= sin 100° (1 / cos 100° + 4).

That simplifies to sin 100° times (sec 100° + 4). But I don't know if that helps. Maybe I need a different approach.

Alternatively, perhaps I can express everything in terms of sine and cosine and then combine the terms. Let me write it as:

(sin 100° / cos 100°) + 4 sin 100°.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Maybe I can get a common denominator for the two terms. The first term has denominator cos 100°, and the second term is just a multiple of sin 100°. So if I write the second term as (4 sin 100° cos 100°)/cos 100°, then both terms will have the same denominator.

Let me try that:

<prompt_2.1> Combine the terms (sin 100° / cos 100°) + 4 sin 100° by getting a common denominator as per the above plan. </prompt_2.1>
-----

Wait, but let me check if there's a better identity or approach here. Sometimes, expressions like a tan θ + b sin θ can be simplified using the identity for tan θ in terms of sin θ, or maybe using auxiliary angles. Let me think. Alternatively, perhaps expressing everything in terms of sine and cosine and then using sum-to-product formulas?

Wait, but since the angle is 100°, which is not a standard angle, maybe we need to find a way to express this as a multiple of a single trigonometric function. Alternatively, maybe using the sine of 100° and tangent of 100°, which is related to sine and cosine. Let me see.

Alternatively, perhaps using the identity for tan θ + k sin θ. Maybe there's a way to combine these terms. Let me consider:

<prompt_2.2> Explore alternative approaches to simplify the expression using trigonometric identities as per the above plan. </prompt_2.2>

Alternatively, maybe using the sine of supplementary angles. Since 100° is 180° - 80°, so sin 100° = sin 80°, and cos 100° = -cos 80°. Let me try substituting that:

<prompt_2.3> Use the identities sin(180° - θ) = sin θ and cos(180° - θ) = -cos θ to simplify the expression as per the above plan. </prompt_2.3>

Alternatively, maybe there's a way to write the original expression as a multiple of a single trigonometric function. Let me see:

Original expression: tan 100° + 4 sin 100°.

Expressed as sin 100° / cos 100° + 4 sin 100° = sin 100° (1 / cos 100° + 4).

Is there a way to combine 1/cos θ + 4 into a single term? Let's see:

<prompt_2.4> Simplify the expression sin 100° (sec 100° + 4) using trigonometric identities as per the above plan. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
= [sin 100° + 4 sin 100° cos 100°] / cos 100°.

Now, factor sin 100° from the numerator:

= sin 100° [1 + 4 cos 100°] / cos 100°.

Hmm, so that's sin 100° times (1 + 4 cos 100°) all over cos 100°. Let me write that as:

sin 100° / cos 100° * [1 + 4 cos 100°] = tan 100° * [1 + 4 cos 100°].
</execution_2.1>

<execution_2.2>
tan θ + 4 sin θ = (sin θ / cos θ) + 4 sin θ = sin θ (1 / cos θ + 4).

But unless 1 / cos θ + 4 can be simplified further, this might not help. Alternatively, perhaps using the identity for tan θ in terms of sin θ and cos θ, but I can't see a direct path here.
</execution_2.2>

<execution_2.3>
tan 100° = tan (180° - 80°) = -tan 80°.

Similarly, sin 100° = sin (180° - 80°) = sin 80°.

So substituting back into the original expression:

tan 100° + 4 sin 100° = -tan 80° + 4 sin 80°.

Hmm, not sure if that helps. Let's see:

-tan 80° + 4 sin 80° = - (sin 80° / cos 80°) + 4 sin 80° = sin 80° (-1 / cos 80° + 4).

Again, similar structure. Maybe this isn't the right path.
</execution_2.3>

<execution_2.4>
1/cos θ + 4 = sec θ + 4. Hmm, perhaps not directly. Alternatively, maybe express 4 as 4 cos θ / cos θ, so:

sec θ + 4 = (1 + 4 cos θ)/cos θ.

Therefore, the entire expression becomes sin θ * (1 + 4 cos θ)/cos θ = [sin θ (1 + 4 cos θ)] / cos θ.

But θ here is 100°, so substituting back:

[sin 100° (1 + 4 cos 100°)] / cos 100°.

Hmm, but is this simpler? Let's compute the numerator:

sin 100° + 4 sin 100° cos 100°.
</execution_2.4>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, but that's just the original expression. So that approach circles back. Maybe there's another identity or method.

Alternatively, perhaps using the sine of sum formula? Let me think.

<prompt_3.1> Consider using the sine of sum formula to simplify the expression as per the above plan. </prompt_3.1>
-----

Based on execution_2.4:
Alternatively, note that sin 100° cos 100° can be expressed using the double-angle identity:

<prompt_3.2> Simplify the expression [sin 100° (1 + 4 cos 100°)] / cos 100° using trigonometric identities as per the above plan. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Let's recall that sin 2θ = 2 sin θ cos θ. So 4 sin θ cos θ is 2 sin 2θ. Wait, in our case, we have 4 sin 100° cos 100°, which would be 2 sin 200°. Let me check:

Yes, sin 2θ = 2 sin θ cos θ, so 4 sin θ cos θ = 2 * 2 sin θ cos θ = 2 sin 2θ. Therefore, 4 sin 100° cos 100° = 2 sin 200°. So substituting back into the expression:

[sin 100° + 4 sin 100° cos 100°] / cos 100° = [sin 100° + 2 sin 200°] / cos 100°.

So now, the numerator is sin 100° + 2 sin 200°. Maybe there's a way to combine these two terms using sum-to-product identities?
</execution_3.1>

<execution_3.2>
sin 2θ = 2 sin θ cos θ. Therefore, sin θ cos θ = (1/2) sin 2θ. So:

4 sin 100° cos 100° = 4*(1/2) sin 200° = 2 sin 200°.

Therefore, the numerator becomes sin 100° + 2 sin 200°. Then, the entire expression is [sin 100° + 2 sin 200°] / cos 100°.

Hmm, maybe that's helpful. Let's see. Then, perhaps we can split this fraction into two terms:

sin 100° / cos 100° + 2 sin 200° / cos 100° = tan 100° + 2 sin 200° / cos 100°.

But the second term is 2 sin 200° / cos 100°. Let's note that sin 200° = sin (180° + 20°) = -sin 20°, so sin 200° = -sin 20°. Therefore:

2 sin 200° / cos 100° = -2 sin 20° / cos 100°.

But cos 100° = cos (180° - 80°) = -cos 80°, so cos 100° = -cos 80°. Therefore:

-2 sin 20° / (-cos 80°) = 2 sin 20° / cos 80°.

So now, the expression becomes tan 100° + 2 sin 20° / cos 80°. But tan 100° is -tan 80°, so:

-tan 80° + 2 sin 20° / cos 80°.

Hmm, let's compute 2 sin 20° / cos 80°. Since cos 80° = sin 10°, because cos(90° - θ) = sin θ. So cos 80° = sin 10°. Therefore:

2 sin 20° / sin 10°. Using the identity sin 20° = 2 sin 10° cos 10°, so:

2*(2 sin 10° cos 10°) / sin 10° = 4 cos 10°.

Therefore, the expression becomes -tan 80° + 4 cos 10°.

But tan 80° = sin 80° / cos 80° = (2 sin 40° cos 40°) / cos 80°, but maybe that's not helpful. Alternatively, tan 80° is a large value, but perhaps this approach isn't leading me anywhere.

Wait, let's recap. Starting from the expression:

tan 100° + 4 sin 100° = [sin 100° (1 + 4 cos 100°)] / cos 100°.

Let me compute the numerical value to check if it's a constant. Maybe the expression simplifies to a constant?

Let me calculate the original expression numerically:

tan 100° + 4 sin 100°.

First, convert 100° to radians: 100° * π/180 ≈ 1.745 radians.

Compute tan(100°): tan(100°) ≈ -5.671.

Compute sin(100°): sin(100°) ≈ 0.9848.

So 4 sin 100° ≈ 4 * 0.9848 ≈ 3.939.

Therefore, tan 100° + 4 sin 100° ≈ -5.671 + 3.939 ≈ -1.732.

Hmm, -1.732 is approximately -√3, since √3 ≈ 1.732. So maybe the simplified form is -√3?
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Let's apply the sum-to-product formula for sin A + sin B:

sin A + sin B = 2 sin[(A + B)/2] cos[(A - B)/2].

<prompt_4.1> Apply the sum-to-product identity to simplify the expression as per the above plan. </prompt_4.1>
-----

Based on execution_3.2:
Let me verify that.

If the expression simplifies to -√3, that would mean:

tan 100° + 4 sin 100° = -√3.

Let me check with approximate values:

<prompt_4.2> Verify if the simplified expression equals -√3 using approximate values as per the above plan. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
But in our case, we have sin 100° + 2 sin 200°. Let me set A = 100° and B = 200°. Then:

sin 100° + sin 200° = 2 sin[(100° + 200°)/2] cos[(100° - 200°)/2] = 2 sin 150° cos(-50°).

But we have sin 100° + 2 sin 200° = sin 100° + sin 200° + sin 200°. So that's sin 100° + sin 200° + sin 200°. Let's first compute sin 100° + sin 200°, then add the remaining sin 200°.

Using the sum-to-product formula for sin 100° + sin 200°:

= 2 sin[(100° + 200°)/2] cos[(100° - 200°)/2]

= 2 sin 150° cos(-50°).

Since sin is an odd function and cos is even, this becomes:

= 2 sin 150° cos 50°.

We know that sin 150° = sin(180° - 30°) = sin 30° = 0.5. So:

= 2 * 0.5 * cos 50° = cos 50°.

Therefore, sin 100° + sin 200° = cos 50°. Then, adding the remaining sin 200°:

Total numerator becomes cos 50° + sin 200°.

But sin 200° = sin(180° + 20°) = -sin 20°, so:

cos 50° - sin 20°.

So the numerator is cos 50° - sin 20°, and the denominator is cos 100°. Therefore, the entire expression becomes:

[cos 50° - sin 20°] / cos 100°.

Hmm, let's see if this can be simplified further.
</execution_4.1>

<execution_4.2>
tan 100° ≈ tan(π - 80°) = -tan 80° ≈ -5.671.

sin 100° ≈ sin(80°) ≈ 0.9848.

So tan 100° + 4 sin 100° ≈ -5.671 + 4*0.9848 ≈ -5.671 + 3.939 ≈ -1.732.

And √3 ≈ 1.732, so -√3 ≈ -1.732.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Let me compute the numerical values to check:

cos 50° ≈ 0.6428,

sin 20° ≈ 0.3420,

so cos 50° - sin 20° ≈ 0.6428 - 0.3420 ≈ 0.3008.

cos 100° ≈ cos(180° - 80°) = -cos 80° ≈ -0.1736.

Therefore, [cos 50° - sin 20°] / cos 100° ≈ 0.3008 / (-0.1736) ≈ -1.732, which is approximately -√3.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Plan_6>
Based on execution_4.2:
Therefore, perhaps the exact value is -√3.

But how to show that algebraically?

Let me see:

Suppose [cos 50° - sin 20°] / cos 100° = -√3.

Multiply both sides by cos 100°:

<prompt_6.1> Verify if the expression equals -√3 algebraically as per the above plan. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
cos 50° - sin 20° = -√3 cos 100°.

Let me compute the left-hand side (LHS):

cos 50° - sin 20°.

Express cos 50° as sin 40°, since cos 50° = sin(90° - 50°) = sin 40°. So:

sin 40° - sin 20°.

Using the identity sin A - sin B = 2 cos[(A + B)/2] sin[(A - B)/2]:

= 2 cos[(40° + 20°)/2] sin[(40° - 20°)/2]

= 2 cos 30° sin 10°.

We know that cos 30° = √3/2, so:

= 2*(√3/2)*sin 10° = √3 sin 10°.

Therefore, LHS is √3 sin 10°, and the right-hand side (RHS) is -√3 cos 100°.

But cos 100° = -cos 80°, so:

RHS = -√3*(-cos 80°) = √3 cos 80°.

Therefore, we have:

√3 sin 10° = √3 cos 80°.

Divide both sides by √3:

sin 10° = cos 80°.

But cos 80° = sin(90° - 80°) = sin 10°, which is true. Therefore, the equality holds.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
Therefore, [cos 50° - sin 20°] / cos 100° = -√3.

Therefore, going back through the steps:

<prompt_7.1> Conclude the simplification of the expression to -√3 as per the above plan. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
tan 100° + 4 sin 100° = [sin 100° (1 + 4 cos 100°)] / cos 100° = [sin 100° + 2 sin 200°] / cos 100° = [cos 50° - sin 20°] / cos 100° = -√3.

Therefore, the simplified form is -√3.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
Wait, but let me check once more. Because when we had [sin 100° + 2 sin 200°] / cos 100°, which became [cos 50° - sin 20°] / cos 100°, and we showed that this equals -√3. Therefore, the original expression simplifies to -√3.

Alternatively, maybe there's a more straightforward way. Let me consider another approach.

<prompt_8.1> Recap the solution and confirm the final simplified form as per the above plan. </prompt_8.1>
</Plan_8>

<Execution_8>
<execution_8.1>
Let me suppose that tan 100° + 4 sin 100° can be written as a constant. Since 100° is a specific angle, perhaps the combination of tan and sin results in a constant value. Given that the approximate value is -√3, and through the identity steps we arrived at -√3, that must be the simplified form.

Alternatively, perhaps there's a geometric interpretation. For example, constructing a triangle or using addition formulas. But given the time I've spent and the consistency in the result, I think the simplified form is indeed -√3.
</execution_8.1>
</Execution_8>

<Final_answer>
Therefore, the simplified form of $\tan 100^\circ + 4 \sin 100^\circ$ is $\boxed{-\sqrt{3}}$.
</Final_answer>
</think>