If `log_(3) sin x-log_(3) cos x-log_(3)(1- tan x)-log_(3)(1+tan x)= -1`, then `tan2x` is equal to (wherever defined)

Options

A. -2
B. `(3)/(2)`
C. `(2)/(3)`
D. 6

Answer

To solve the equation \( \log_3 \sin x - \log_3 \cos x - \log_3(1 - \tan x) - \log_3(1 + \tan x) = -1 \), we will follow these steps:

### Step 1: Combine the logarithmic terms
Using the property of logarithms that states \( \log_a b - \log_a c = \log_a \left(\frac{b}{c}\right) \), we can combine the logarithmic terms:

\[ \log_3 \left(\frac{\sin x}{\cos x}\right) - \log_3 \left((1 - \tan x)(1 + \tan x)\right) = -1 \]

### Step 2: Simplify the expression
Using the identity \( \tan x = \frac{\sin x}{\cos x} \), we rewrite \( \frac{\sin x}{\cos x} \) as \( \tan x \):

\[ \log_3 \tan x - \log_3 \left(1 - \tan^2 x\right) = -1 \]

### Step 3: Combine the logs again
Now, we can combine the logs again:

\[ \log_3 \left(\frac{\tan x}{1 - \tan^2 x}\right) = -1 \]

### Step 4: Convert the logarithmic equation to exponential form
This means:

\[ \frac{\tan x}{1 - \tan^2 x} = 3^{-1} = \frac{1}{3} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives:

\[ \tan x = \frac{1}{3}(1 - \tan^2 x) \]

### Step 6: Rearranging the equation
Rearranging the equation leads to:

\[ \tan x + \frac{1}{3} \tan^2 x = \frac{1}{3} \]

Multiplying through by 3 to eliminate the fraction:

\[ 3\tan x + \tan^2 x = 1 \]

### Step 7: Rearranging into standard quadratic form
Rearranging gives:

\[ \tan^2 x + 3\tan x - 1 = 0 \]

### Step 8: Solving the quadratic equation
Using the quadratic formula \( \tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1, b = 3, c = -1 \):

\[ \tan x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ \tan x = \frac{-3 \pm \sqrt{9 + 4}}{2} \]
\[ \tan x = \frac{-3 \pm \sqrt{13}}{2} \]

### Step 9: Finding \( \tan 2x \)
Using the double angle formula for tangent:

\[ \tan 2x = \frac{2\tan x}{1 - \tan^2 x} \]

Substituting \( \tan x = \frac{-3 + \sqrt{13}}{2} \) (taking the positive root for simplicity):

1. Calculate \( \tan^2 x \):
\[ \tan^2 x = \left(\frac{-3 + \sqrt{13}}{2}\right)^2 = \frac{(-3 + \sqrt{13})^2}{4} \]

2. Substitute into the formula for \( \tan 2x \):
\[ \tan 2x = \frac{2 \cdot \frac{-3 + \sqrt{13}}{2}}{1 - \left(\frac{-3 + \sqrt{13}}{2}\right)^2} \]

After calculating, we find:

\[ \tan 2x = \frac{2(-3 + \sqrt{13})}{1 - \frac{(-3 + \sqrt{13})^2}{4}} \]

This simplifies to:

\[ \tan 2x = \frac{2(-3 + \sqrt{13})}{\frac{4 - (-3 + \sqrt{13})^2}{4}} = \frac{8(-3 + \sqrt{13})}{4 - (-3 + \sqrt{13})^2} \]

### Final Answer
Thus, the value of \( \tan 2x \) is \( \frac{2}{3} \).

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