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

First, let's recall a key relationship between inverse tangent and inverse cotangent:

$\cot^{-1}(z) = \tan^{-1}\left(\dfrac{1}{z}\right)$

This relationship exists because cotangent is the reciprocal of tangent. Geometrically, if an angle has a tangent value of z, then the same angle has a cotangent value of 1/z.

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Using this relationship, let's rewrite the right side of our equation:

$\cot^{-1}\left(\dfrac{3}{3^x+1}\right) = \tan^{-1}\left(\dfrac{3^x+1}{3}\right)$

Our equation becomes:

$\tan^{-1}\left(\dfrac{2}{3^{-x}+1}\right) = \tan^{-1}\left(\dfrac{3^x+1}{3}\right)$

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For two inverse tangent expressions to be equal, their arguments must be equal (since inverse tangent is a one-to-one function in its principal domain):

$\dfrac{2}{3^{-x}+1} = \dfrac{3^x+1}{3}$

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Cross-multiplying to solve:

$2 \cdot 3 = (3^{-x}+1)(3^x+1)$

$6 = 3^{-x} \cdot 3^x + 3^{-x} + 3^x + 1$

$6 = 1 + 3^{-x} + 3^x + 1$

$4 = 3^{-x} + 3^x$

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Remember that $3^{-x} = \dfrac{1}{3^x}$, so we can rewrite:

$4 = \dfrac{1}{3^x} + 3^x$

$4 \cdot 3^x = 1 + (3^x)^2$

$(3^x)^2 - 4(3^x) + 1 = 0$

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Let's substitute $y = 3^x$ to make this easier to solve:

$y^2 - 4y + 1 = 0$

Using the quadratic formula:

$y = \dfrac{4 \pm \sqrt{16-4}}{2} = \dfrac{4 \pm \sqrt{12}}{2} = \dfrac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$

So we have two solutions:

$y = 3^x = 2 + \sqrt{3}$

$y = 3^x = 2 - \sqrt{3}$

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Now let's find the x-values by taking logarithm base 3:

For the first solution:

$3^x = 2 + \sqrt{3}$

$x = \log_3(2 + \sqrt{3})$

Since $2 + \sqrt{3} \approx 3.732 > 1$, this value of x is positive.

For the second solution:

$3^x = 2 - \sqrt{3}$

$x = \log_3(2 - \sqrt{3})$

Since $2 - \sqrt{3} \approx 0.268 < 1$, this value of x is negative.

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A quick conceptual check: When we have an equation with exponential terms like $3^x$ and $3^{-x}$, it's common to get solutions in pairs - one positive and one negative. This happens because of the symmetrical nature of the equation. We can see this symmetry in the final form $4 = 3^{-x} + 3^x$ where replacing x with -x gives us the same equation.

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Therefore, there is one positive real value $x = \log_3(2 + \sqrt{3})$ and one negative real value $x = \log_3(2 - \sqrt{3})$ satisfying the equation.

The correct option is: There is one positive and one negative real value of x satisfying the above equation.