If cosec `theta` = 3x and cot `theta = (3)/(x)` , then find the value of `(x^(2) - (1)/(x^(2)))` .

Answer

To solve the problem, we start with the given equations:

1. \( \csc \theta = 3x \)
2. \( \cot \theta = \frac{3}{x} \)

We need to find the value of \( x^2 - \frac{1}{x^2} \).

### Step 1: Express \( \sin \theta \) and \( \cos \theta \)

From the definition of cosecant and cotangent, we know that:

\[ \csc \theta = \frac{1}{\sin \theta} \implies \sin \theta = \frac{1}{3x} \]

\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \implies \cos \theta = \cot \theta \cdot \sin \theta = \frac{3}{x} \cdot \frac{1}{3x} = \frac{1}{x^2} \]

### Step 2: Use the Pythagorean identity

We know that:

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

Substituting the values of \( \sin \theta \) and \( \cos \theta \):

\[ \left(\frac{1}{3x}\right)^2 + \left(\frac{1}{x^2}\right)^2 = 1 \]

This simplifies to:

\[ \frac{1}{9x^2} + \frac{1}{x^4} = 1 \]

### Step 3: Clear the denominators

To eliminate the fractions, multiply through by \( 9x^4 \):

\[ 9x^4 \cdot \frac{1}{9x^2} + 9x^4 \cdot \frac{1}{x^4} = 9x^4 \]

This simplifies to:

\[ x^2 + 9 = 9x^4 \]

### Step 4: Rearrange the equation

Rearranging gives us:

\[ 9x^4 - x^2 - 9 = 0 \]

### Step 5: Substitute \( y = x^2 \)

Let \( y = x^2 \). Then the equation becomes:

\[ 9y^2 - y - 9 = 0 \]

### Step 6: Solve the quadratic equation

Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 9, b = -1, c = -9 \):

\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 9 \cdot (-9)}}{2 \cdot 9} \]

Calculating the discriminant:

\[ = \frac{1 \pm \sqrt{1 + 324}}{18} = \frac{1 \pm \sqrt{325}}{18} \]

### Step 7: Find \( x^2 - \frac{1}{x^2} \)

Now we need to find \( x^2 - \frac{1}{x^2} \):

Using \( y = x^2 \):

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

Substituting \( y \) into the expression:

\[ = \frac{y^2 - 1}{y} \]

### Final Step: Substitute back the value of \( y \)

Using the value of \( y \) from the quadratic solution, we can calculate \( x^2 - \frac{1}{x^2} \).