The correct option is: There is one positive and one negative real value of x satisfying the abov eequation

Step-by-Step Explanation:

From the given image, we have the equation:
$${\tan }^{-1}\left(\frac{2}{3^{-x}+1}\right)={\cot }^{-1}\left(\frac{3}{3^x+1}\right)$$

First, recall the relationship between the inverse trigonometric functions: for any y > 0,
$${\cot }^{-1}\left(y\right)={\tan }^{-1}\left(\frac{1}{y}\right)$$

Since 3^x > 0 for all real numbers x, the argument of the cotangent function is always positive:
$$\frac{3}{3^x+1}>0$$

We can rewrite the right-hand side of our equation as:
$${\cot }^{-1}\left(\frac{3}{3^x+1}\right)={\tan }^{-1}\left(\frac{3^x+1}{3}\right)$$

Now, equating the arguments of the arctangent functions from both sides of the equation yields:
$$\frac{2}{3^{-x}+1}=\frac{3^x+1}{3}$$

We simplify the term 3^-x + 1 on the left side of the equation:
$$3^{-x}+1=\frac{1}{3^x}+1=\frac{1+3^x}{3^x}$$

Substituting this back into the equation, we get:
$$\frac{2·3^x}{3^x+1}=\frac{3^x+1}{3}$$

Cross-multiplying the denominators, we obtain:
$$6·3^x={\left(3^x+1\right)}^2$$

Let y = 3^x. Since x must be a real number, y must be strictly positive (y > 0). Substituting y into the equation gives:
$$6y={\left(y+1\right)}^2$$

Expanding the squared term and rearranging the equation into standard quadratic form:
$$6y=y^2+2y+1$$
$$y^2-4y+1=0$$

Applying the quadratic formula to solve for y:
$$y=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4·1·1}}{2·1}$$
$$y=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}$$

Since √3 ≈ 1.732, both solutions for y are positive:
1. y₁ = 2 + √3 ≈ 3.732 > 0
2. y₂ = 2 - √3 ≈ 0.268 > 0

Now we solve for the corresponding real values of x using 3^x = y:
For y₁ = 2 + √3:
$$3^x=2+\sqrt{3}$$
Since 2 + √3 > 1, the exponent x = log₃(2 + √3) is positive (x > 0).

For y₂ = 2 - √3:
$$3^x=2-\sqrt{3}$$
Since 0 < 2 - √3 < 1, the exponent x = log₃(2 - √3) is negative (x < 0).

Consequently, there is exactly one positive and one negative real value of x that satisfies the equation.