Let $f$ be a twice differentiable function such that $f(x)=\int_{0}^{x}\tan(t-x)dt-\int_{0}^{x}f(t)\tan t\,dt$, $x \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. Then $f''\left(\dfrac{\pi}{6}\right)+12f'\left(-\dfrac{\pi}{6}\right)+f\left(\dfrac{\pi}{6}\right)$ is equal to ______

Given the function:

$f(x) = \int_{0}^{x} \tan(t-x) dt - \int_{0}^{x} f(t) \tan t \, dt$

First, we simplify the first integral. Let $u = x - t$, then $du = -dt$. When $t = 0$, $u = x$; when $t = x$, $u = 0$.

$\int_{0}^{x} \tan(t-x) dt = \int_{x}^{0} \tan(-u) (-du) = \int_{x}^{0} \tan u \, du = -\int_{0}^{x} \tan u \, du = -[\ln(\sec u)]_{0}^{x} = -\ln(\sec x) = \ln(\cos x)$

So, the equation becomes:

$f(x) = \ln(\cos x) - \int_{0}^{x} f(t) \tan t \, dt$

Differentiating both sides with respect to $x$ using the Leibniz rule:

$f'(x) = \dfrac{-\sin x}{\cos x} - f(x) \tan x$

$f'(x) + f(x) \tan x = -\tan x$

This is a linear first-order differential equation. The integrating factor (IF) is:

$\text{IF} = e^{\int \tan x \, dx} = e^{\ln(\sec x)} = \sec x$

Multiplying the differential equation by $\sec x$:

$f'(x) \sec x + f(x) \sec x \tan x = -\sec x \tan x$

$\dfrac{d}{dx} (f(x) \sec x) = -\sec x \tan x$

Integrating both sides with respect to $x$:

$f(x) \sec x = -\sec x + C$

To find $C$, we use the initial condition. From $f(x) = \ln(\cos x) - \int_{0}^{x} f(t) \tan t \, dt$, substituting $x = 0$ gives $f(0) = \ln(1) - 0 = 0$.

$0 \cdot \sec 0 = -\sec 0 + C \Rightarrow 0 = -1 + C \Rightarrow C = 1$

Thus, $f(x) \sec x = 1 - \sec x$

$f(x) = \cos x - 1$

Now, we find the required derivatives:

$f'(x) = -\sin x$

$f''(x) = -\cos x$

Evaluating these at the given points:

$f\left(\dfrac{\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}\right) - 1 = \dfrac{\sqrt{3}}{2} - 1$

$f'\left(-\dfrac{\pi}{6}\right) = -\sin\left(-\dfrac{\pi}{6}\right) = \sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$

$f''\left(\dfrac{\pi}{6}\right) = -\cos\left(\dfrac{\pi}{6}\right) = -\dfrac{\sqrt{3}}{2}$

Finally, substituting these values into the required expression:

$f''\left(\dfrac{\pi}{6}\right) + 12f'\left(-\dfrac{\pi}{6}\right) + f\left(\dfrac{\pi}{6}\right) = -\dfrac{\sqrt{3}}{2} + 12\left(\dfrac{1}{2}\right) + \left(\dfrac{\sqrt{3}}{2} - 1\right)$

$= -\dfrac{\sqrt{3}}{2} + 6 + \dfrac{\sqrt{3}}{2} - 1 = 5$

Answer: $5$