To solve the problem, we need to find 
tan
⁡
(
𝑥
+
𝑦
)
tan
⁡
(
𝑥
−
𝑦
)
tan(x−y)
tan(x+y)
​
  given that 
sin
⁡
(
2
𝑥
)
=
101
sin
⁡
(
2
𝑦
)
sin(2x)=101sin(2y).



To solve the problem of finding the ratio \(\frac{\tan(x+y)}{\tan(x-y)}\), given that \(\sin(2x) = 101\sin(2y)\), we can follow these steps:

1. **Express \(\sin(2x)\) and \(\sin(2y)\) in terms of \(\tan\):**

   Recall that:
   \[
   \sin(2x) = 2\sin(x)\cos(x) = \frac{2\tan(x)}{1+\tan^2(x)}
   \]
   and similarly,
   \[
   \sin(2y) = 2\sin(y)\cos(y) = \frac{2\tan(y)}{1+\tan^2(y)}
   \]

2. **Substitute into the given equation:**

   Given \(\sin(2x) = 101\sin(2y)\), we substitute the expressions for \(\sin(2x)\) and \(\sin(2y)\):
   \[
   \frac{2\tan(x)}{1+\tan^2(x)} = 101 \cdot \frac{2\tan(y)}{1+\tan^2(y)}
   \]

   Dividing both sides by 2, we get:
   \[
   \frac{\tan(x)}{1+\tan^2(x)} = 101 \cdot \frac{\tan(y)}{1+\tan^2(y)}
   \]

3. **Express \(\tan(x+y)\) and \(\tan(x-y)\):**

   Using the tangent addition and subtraction formulas:
   \[
   \tan(x+y) = \frac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)}
   \]
   \[
   \tan(x-y) = \frac{\tan(x) - \tan(y)}{1 + \tan(x)\tan(y)}
   \]

4. **Find \(\frac{\tan(x+y)}{\tan(x-y)}\):**

   \[
   \frac{\tan(x+y)}{\tan(x-y)} = \frac{\frac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)}}{\frac{\tan(x) - \tan(y)}{1 + \tan(x)\tan(y)}}
   \]

   Simplifying this expression, we get:
   \[
   \frac{\tan(x+y)}{\tan(x-y)} = \frac{(\tan(x) + \tan(y))(1 + \tan(x)\tan(y))}{(\tan(x) - \tan(y))(1 - \tan(x)\tan(y))}
   \]

5. **Relate \(\tan(x)\) and \(\tan(y)\):**

   Using \(\frac{\tan(x)}{1+\tan^2(x)} = 101 \cdot \frac{\tan(y)}{1+\tan^2(y)}\), let \(\tan(x) = t\) and \(\tan(y) = u\). Then:
   \[
   \frac{t}{1+t^2} = 101 \cdot \frac{u}{1+u^2}
   \]

   Solving this equation for \(t\) in terms of \(u\), we get:
   \[
   t(1+u^2) = 101u(1+t^2)
   \]
   This equation can be used to find the relationship between \(\tan(x)\) and \(\tan(y)\).

Given these relationships, one can plug into the expression for \(\frac{\tan(x+y)}{\tan(x-y)}\) and simplify further, depending on the values or the relationships found.

Would you like more details on solving the equation for \(t\) and \(u\), or is there a specific part of the process you want to focus on?

### Relative Questions:
1. How can we solve \(\frac{t}{1+t^2} = 101 \cdot \frac{u}{1+u^2}\) explicitly for \(t\) in terms of \(u\)?
2. What other trigonometric identities might be useful in solving this problem?
3. Can the relationship between \(\tan(x)\) and \(\tan(y)\) be used to simplify the expression for \(\tan(x+y)\) and \(\tan(x-y)\)?
4. Are there any special angles or values of \(x\) and \(y\) that satisfy the given condition?
5. How can we interpret the equation geometrically or graphically?

### Tip:
When dealing with trigonometric identities and equations, it's often helpful to express everything in terms of a single trigonometric function (like \(\tan\)) to simplify the algebra involved.

Learn how to find the ratio tan(x+y) / tan(x-y) using trigonometric identities and equations. Given sin(2x) = 101sin(2y), this problem involves applying tangent addition and subtraction formulas. Suitable for advanced high school and college-level students.

Math Problem Statement

To solve the problem, we need to find tan ⁡ ( 𝑥 + 𝑦 ) tan ⁡ ( 𝑥 − 𝑦 ) tan(x−y) tan(x+y) given that sin ⁡ ( 2 𝑥 )

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