The value of `(2 sinx)/(sin 3x)+(tanx)/(tan3x)` ____________

Explanation

To solve the expression \(\frac{2 \sin x}{\sin 3x} + \frac{\tan x}{\tan 3x}\), we will follow these steps:

### Step 1: Rewrite \(\sin 3x\) and \(\tan 3x\)
Using the triple angle formulas:
- \(\sin 3x = 3 \sin x - 4 \sin^3 x\)
- \(\tan 3x = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x}\)

### Step 2: Substitute the values into the expression
Now we substitute these values into the expression:
\[ \frac{2 \sin x}{3 \sin x - 4 \sin^3 x} + \frac{\tan x}{\frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x}} \]

### Step 3: Simplify the second term
The second term can be simplified:
\[ \frac{\tan x (1 - 3 \tan^2 x)}{3 \tan x - \tan^3 x} \]

### Step 4: Rewrite the expression
Now we can rewrite the entire expression:
\[ \frac{2 \sin x}{3 \sin x - 4 \sin^3 x} + \frac{\tan x (1 - 3 \tan^2 x)}{3 \tan x - \tan^3 x} \]

### Step 5: Factor out common terms
Notice that \(\tan x = \frac{\sin x}{\cos x}\), so we can rewrite the second term:
\[ \frac{\frac{\sin x}{\cos x} (1 - 3 \frac{\sin^2 x}{\cos^2 x})}{3 \frac{\sin x}{\cos x} - \left(\frac{\sin^3 x}{\cos^3 x}\right)} \]

### Step 6: Cancel out \(\sin x\) and \(\tan x\)
Both terms have \(\sin x\) in the numerator, which can be canceled out:
\[ \frac{2}{3 - 4 \sin^2 x} + \frac{(1 - 3 \sin^2 x)}{3 - 3 \sin^2 x} \]

### Step 7: Combine the fractions
Now we can combine the two fractions:
\[ \frac{2(3 - 3 \sin^2 x) + (1 - 3 \sin^2 x)(3 - 4 \sin^2 x)}{(3 - 4 \sin^2 x)(3 - 3 \sin^2 x)} \]

### Step 8: Simplify the numerator
Expanding the numerator:
\[ 6 - 6 \sin^2 x + 3 - 12 \sin^2 x + 4 \sin^4 x - 3 \sin^2 x + 12 \sin^4 x \]
Combine like terms:
\[ 9 - 21 \sin^2 x + 16 \sin^4 x \]

### Step 9: Final simplification
The denominator simplifies to:
\[ (3 - 4 \sin^2 x)(3 - 3 \sin^2 x) \]
Now, we can see that both the numerator and denominator have common factors that can be canceled, leading to:
\[ 1 \]

### Final Answer
Thus, the value of the expression \(\frac{2 \sin x}{\sin 3x} + \frac{\tan x}{\tan 3x}\) is:
\[ \boxed{1} \]