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 can follow these steps:

### Step 1: Combine the logarithmic expressions
Using the properties of logarithms, we can combine the logarithmic terms:
\[ \log_{3} \sin x - \log_{3} \cos x = \log_{3} \left(\frac{\sin x}{\cos x}\right) = \log_{3} (\tan x) \]
And for the other terms:
\[ -\log_{3}(1 - \tan x) - \log_{3}(1 + \tan x) = -\log_{3}((1 - \tan x)(1 + \tan x)) = -\log_{3}(1 - \tan^2 x) \]
Thus, we can rewrite the equation as:
\[ \log_{3} (\tan x) - \log_{3}(1 - \tan^2 x) = -1 \]

### Step 2: Use the properties of logarithms to simplify
Now, we can combine these logarithms:
\[ \log_{3} \left(\frac{\tan x}{1 - \tan^2 x}\right) = -1 \]

### Step 3: Convert from logarithmic to exponential form
Using the definition of logarithms, we can convert this to exponential form:
\[ \frac{\tan x}{1 - \tan^2 x} = 3^{-1} = \frac{1}{3} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ \tan x = \frac{1}{3} (1 - \tan^2 x) \]

### Step 5: Rearranging the equation
Rearranging this 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 6: Rearranging into standard quadratic form
Rearranging gives us:
\[ \tan^2 x + 3 \tan x - 1 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( \tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 3, c = -1 \):
\[ \tan x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 4}}{2} = \frac{-3 \pm \sqrt{13}}{2} \]

### Step 8: Find \( \tan 2x \)
Using the double angle formula for tangent:
\[ \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \]
Substituting \( \tan x = \frac{-3 + \sqrt{13}}{2} \) or \( \tan x = \frac{-3 - \sqrt{13}}{2} \) into the formula will yield the value of \( \tan 2x \).

Calculating \( \tan^2 x \):
\[ \tan^2 x = \left(\frac{-3 + \sqrt{13}}{2}\right)^2 = \frac{(-3 + \sqrt{13})^2}{4} = \frac{9 - 6\sqrt{13} + 13}{4} = \frac{22 - 6\sqrt{13}}{4} \]
Now substituting back into \( \tan 2x \):
\[ \tan 2x = \frac{2 \cdot \frac{-3 + \sqrt{13}}{2}}{1 - \frac{22 - 6\sqrt{13}}{4}} = \frac{-3 + \sqrt{13}}{1 - \frac{22 - 6\sqrt{13}}{4}} \]
Simplifying this will yield \( \tan 2x = \frac{2}{3} \).

### Final Answer
Thus, \( \tan 2x = \frac{2}{3} \).

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