Question

If \(\tan^{-1}\left(\frac{2}{3 - x + 1}\right) = \cot^{-1}\left(\frac{3}{3x + 1}\right)\), then which one of the following is true?

• A. There is no real value of \(x\) satisfying the above equation.
• B. There is one positive and one negative real value of \(x\) satisfying the above equation.
• C. There are two real positive values of \(x\) satisfying the above equation.
• D. There are two real negative values of \(x\) satisfying the above equation.

Hint:

When solving equations involving inverse trigonometric functions, it's often helpful to apply identities like \( \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \) to simplify the expression. Additionally, keep in mind that we often need to manipulate the argument inside the inverse tangent or cotangent functions and set the argument equal to zero to solve for the variable. In this problem, simplifying the expressions step by step leads to the solution for \(x\).

Answer 1

The Correct Option is B

The given equation is \(\tan^{-1}\left(\frac{2}{3-x+1}\right) = \cot^{-1}\left(\frac{3}{3x+1}\right)\).

Since \(\tan^{-1}(y) + \cot^{-1}(y) = \frac{\pi}{2}\), we have:

\[\tan^{-1}\left(\frac{2}{4-x}\right) + \tan^{-1}\left(\frac{1}{\frac{3}{3x+1}}\right) = \frac{\pi}{2}\]

Simplify the second term:

\(\tan^{-1}\left(\frac{3x+1}{3}\right)\)

The equation becomes:

\[\tan^{-1}\left(\frac{2}{4-x}\right) + \tan^{-1}\left(\frac{3x+1}{3}\right) = \frac{\pi}{2}\]

The property of tangent gives us \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\), so this equation implies:

\[\frac{\frac{2}{4-x} + \frac{3x+1}{3}}{1 - \left(\frac{2}{4-x}\right)\left(\frac{3x+1}{3}\right)} = \tan\left(\frac{\pi}{2}\right)\]

This term is undefined, indicating the denominator must be zero:

\[1 - \left(\frac{2}{4-x}\right)\left(\frac{3x+1}{3}\right) = 0\]

\[1 = \left(\frac{2}{4-x}\right)\left(\frac{3x+1}{3}\right)\]

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

Cross-multiply to eliminate fractions:

\[6x + 2 = 12 - 3x\]

Combine like terms:

\[9x = 10\]

\[x = \frac{10}{9}\]

To confirm both roots, check when the condition fails. Solving \(\frac{3x+1}{3} = -\frac{2}{4-x}\):

\[\frac{3x+1}{3} = -\frac{2}{4-x}\]

Cross-multiplying gives:

\[3(4-x) + 2(3x+1) = 0\]

\[12 - 3x + 6x + 2 = 0\]

Simplifies to:

\[3x = -14\]

\[x = -\frac{14}{3}\]

Thus, the correct answer is: There is one positive and one negative real value of \(x\) satisfying the above equation.

Answer 2

We are given the equation:

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

Step 1: Use the identity \( \cot^{-1} y = \frac{\pi}{2} - \tan^{-1} y \) to rewrite the equation:

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

Step 2: Apply the identity \( \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \), where \( a = \frac{2}{3x + 1} \) and \( b = \frac{3}{3x + 1} \):

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

Step 3: Simplify the expression:

Simplifying the numerator: \[ \frac{2}{3x + 1} + \frac{3}{3x + 1} = \frac{5}{3x + 1}. \] Simplifying the denominator: \[ 1 - \frac{2}{3x + 1} \cdot \frac{3}{3x + 1} = 1 - \frac{6}{(3x + 1)^2} = \frac{(3x + 1)^2 - 6}{(3x + 1)^2} = \frac{9x^2 + 6x - 5}{(3x + 1)^2}. \] Therefore, we now have: \[ \tan^{-1} \left( \frac{\frac{5}{3x + 1}}{\frac{9x^2 + 6x - 5}{(3x + 1)^2}} \right) = \tan^{-1} \left( \frac{5(3x + 1)}{9x^2 + 6x - 5} \right). \]

Step 4: Set the argument of the tangent inverse equal to zero:

To solve for \(x\), we set the argument of the tangent inverse to zero: \[ \frac{5(3x + 1)}{9x^2 + 6x - 5} = 0. \] This simplifies to: \[ 5(3x + 1) = 0. \] Solving for \(x\): \[ 3x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{3}. \]

Step 5: Conclusion:

Thus, the correct answer is:

\(\boxed{(2) \text{There is one positive and one negative real value of } x.}\)