The sum of possible values of x for tan⁻¹(x + 1) + cot⁻¹(1/(x - 1)) = tan⁻¹(8/31) is (A) -32/4 (B) -

Q: Wait, why do we get two answers from the quadratic equation, but only one is correct? Did I make a calculation error?

You didn't make a calculation error! This is a classic JEE trap. When we apply the tangent function to both sides of an equation like $\tan^{-1}(A) + \cot^{-1}(B) = C$, we are essentially solving $\tan(\text{LHS}) = \tan(\text{RHS})$. The tangent function is periodic, and the equation $\tan(A+B) = \tan(C)$ is satisfied not just when $A+B = C$, but also when $A+B = C + n\pi$. Thus, applying $\tan$ can introduce extraneous roots. Always verify your solutions by plugging them back into the original equation.

Q: Why can't I just use $\cot^{-1}(1/z) = \tan^{-1}(z)$ blindly?

That identity is only true for $z > 0$. If $z < 0$, $\cot^{-1}(1/z)$ actually equals $\pi + \tan^{-1}(z)$. Because the value of $x$ can be negative, the argument $1/(x-1)$ can be negative, which changes the branch of the inverse trigonometric function. Ignoring this domain restriction is the most common way students lose marks in these problems.

Q: How do I know when to reject a root without calculating the exact value?

You can use estimation. For $x = 1/4$, the LHS becomes $\tan^{-1}(1.25) + \cot^{-1}(-4/3)$. Since $\cot^{-1}(-4/3)$ is in the second quadrant (greater than $\pi/2$), the sum is clearly greater than $\pi/2$. However, the RHS is $\tan^{-1}(8/31)$, which is a small positive angle less than $\pi/4$. Since LHS > RHS, $x = 1/4$ cannot be a solution. Estimation is a powerful tool to save time!

Question

The sum of possible values of $x$ for $	an^{-1}(x + 1) + \cot^{-1} \left( \frac{1}{x-1} ight) = 	an^{-1} \left( \frac{8}{31} ight)$ is

Options:

(A) $-\frac{32}{4}$

(B) $-\frac{31}{4}$

(C) $-\frac{30}{4}$

(D) $-\frac{33}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow travelers of the JEE path. Today, we are not just solving an equation; we are peeling back the layers of a beautiful, deceptive problem.

At first glance, the equation $	an^{-1}(x + 1) + \cot^{-1} \left( \frac{1}{x-1} ight) = 	an^{-1} \left( \frac{8}{31} ight)$ looks like a standard algebraic exercise. But beneath the surface lies a trap that has ensnared many brilliant minds.

### The Bridge to Algebra

Our first instinct is to bridge the gap between the transcendental world of inverse trigonometry and the concrete world of algebra. We apply the tangent function to both sides to simplify the expression.

We define $A = 	an^{-1}(x+1)$ and $B = \cot^{-1} \left( \frac{1}{x-1} ight)$. By applying the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$, we transform the equation into:

$$ \frac{(x+1) + (x-1)}{1 - (x+1)(x-1)} = \frac{8}{31} $$

Notice the elegance here: the numerator simplifies to $2x$, and the denominator becomes $1 - (x^2 - 1) = 2 - x^2$. We are left with the clean, manageable quadratic:

$$ \frac{2x}{2 - x^2} = \frac{8}{31} $$

### The Quadratic Dance

Cross-multiplying gives us $31(2x) = 8(2 - x^2)$, which simplifies to $62x = 16 - 8x^2$. Rearranging, we get $8x^2 + 62x - 16 = 0$.

Dividing by $2$, we arrive at $4x^2 + 31x - 8 = 0$. Factoring this is a joy: $4x^2 + 32x - x - 8 = 0$, leading to $(4x - 1)(x + 8) = 0$.

We have two candidates: $x = \frac{1}{4}$ and $x = -8$. But here, the story takes a turn.

### The JEE Trap: The Moment of Truth

In the world of JEE, finding the roots is only half the battle. We must verify them, as applying the tangent function potentially introduced extraneous roots.

Let us test $x = \frac{1}{4}$. The LHS becomes $	an^{-1}(1.25) + \cot^{-1}(-4/3)$. Since $\cot^{-1}(-4/3) = \pi - \cot^{-1}(4/3)$, the LHS is clearly greater than $\frac{\pi}{2}$.

However, the RHS is $	an^{-1}(8/31)$, which is a small positive angle. They cannot be equal. Thus, $x = \frac{1}{4}$ is rejected.

Now, let us test $x = -8$. The LHS becomes $	an^{-1}(-7) + \cot^{-1}(-1/9)$. Using the property $\cot^{-1}(-y) = \pi - \cot^{-1}(y)$, we find that the expression simplifies perfectly to $	an^{-1}(8/31)$.

The cancellation is not just algebraic; it is a geometric necessity. The only valid solution is **$x = -8$**. We have navigated the trap, verified our path, and arrived at the truth.

Animated Solution for Mathematics - Inverse Trigonometric Functions: The sum of possible values of $x$ for $tan^{- 1} (x + 1) + cot^{- 1} (\frac{1}{x - 1}) = tan^{- 1} (\frac{8}{31})$ is

Select Answer:

Visualized Solution

Analyze the Equation

Apply Tangent to Both Sides

Substitute into the Formula

Simplify the Rational Expression

Form the Quadratic Equation

Solve for $x$

The JEE Trap: Extraneous Roots

Verify $x = 1/4$

Verify $x = - 8$

Final Conclusion

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

The Bridge to Algebra

The Quadratic Dance

The JEE Trap

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