We are given the equation defining the function $f(x)$: $f(x) = \int_{0}^{x} \tan(t - x)dt - \int_{0}^{x} f(t)\tan t dt$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. We need to find the value of $f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$.

Simplifying the Integral Equation

First, simplify the integral $\int_{0}^{x} \tan(t - x)dt$. Let $u = t - x$. Then $du = dt$. The limits change from $t=0 \to u=-x$ and $t=x \to u=0$. The integral becomes $\int_{-x}^{0} \tan(u)du$. The antiderivative of $\tan(u)$ is $-\ln|\cos u|$. So, $\int_{-x}^{0} \tan(u)du = [-\ln|\cos u|]_{-x}^{0} = (-\ln|\cos 0|) - (-\ln|\cos(-x)|) = -\ln(1) + \ln|\cos x| = \ln(\cos x)$ (since $\cos x > 0$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$). Thus, the equation simplifies to: $f(x) = \ln(\cos x) - \int_{0}^{x} f(t)\tan t dt$.

Finding the Differential Equation

Differentiate both sides with respect to $x$ using the Fundamental Theorem of Calculus:

$f'(x) = \frac{d}{dx}(\ln(\cos x)) - \frac{d}{dx}\left(\int_{0}^{x} f(t)\tan t dt\right)$

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

$f'(x) = -\tan x - f(x)\tan x = -\tan x (1 + f(x)) \quad (*)$

Differentiate again to find $f''(x)$:

$f''(x) = \frac{d}{dx}(-\tan x (1 + f(x)))$

Using the product rule:

$f''(x) = (-\sec^2 x)(1 + f(x)) - \tan x (f'(x))$

Substitute $f'(x)$ from equation $(*)$:

$f''(x) = -\sec^2 x (1 + f(x)) - \tan x (-\tan x (1 + f(x)))$

$f''(x) = -\sec^2 x (1 + f(x)) + \tan^2 x (1 + f(x))$

$f''(x) = (\tan^2 x - \sec^2 x)(1 + f(x))$

Using the identity $\sec^2 x - \tan^2 x = 1$, we have $\tan^2 x - \sec^2 x = -1$.

$f''(x) = (-1)(1 + f(x)) = -1 - f(x)$

Rearranging gives the differential equation: $f''(x) + f(x) = -1$

Solving the Differential Equation

The general solution to $f''(x) + f(x) = -1$ is $f(x) = A\cos x + B\sin x - 1$. We need to find the constants $A$ and $B$. First, find $f(0)$. From the original integral equation, setting $x=0$: $f(0) = \int_{0}^{0} \tan(t)dt - \int_{0}^{0} f(t)\tan t dt = 0 - 0 = 0$. Substitute $f(0)=0$ into the general solution: $0 = A\cos 0 + B\sin 0 - 1 = A(1) + B(0) - 1 = A - 1$. So, $A = 1$. The solution becomes $f(x) = \cos x + B\sin x - 1$. Now, use the relationship $f'(x) = -\tan x (1 + f(x))$. The derivative is $f'(x) = -\sin x + B\cos x$. Substitute $f(x)$ and $f'(x)$ into the relation: $-\sin x + B\cos x = -\tan x (1 + (\cos x + B\sin x - 1))$ $-\sin x + B\cos x = -\tan x (\cos x + B\sin x)$ $-\sin x + B\cos x = -\frac{\sin x}{\cos x}(\cos x + B\sin x)$ Multiply by $\cos x$: $(-\sin x + B\cos x)\cos x = -\sin x (\cos x + B\sin x)$ $-\sin x \cos x + B\cos^2 x = -\sin x \cos x - B\sin^2 x$ $B\cos^2 x = -B\sin^2 x$ $B(\cos^2 x + \sin^2 x) = 0$ $B(1) = 0$, which implies $B = 0$. Therefore, the function is $f(x) = \cos x - 1$.

Calculating the Final Value

We need to evaluate $f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$. Using $f(x) = \cos x - 1$: $f'(x) = -\sin x$ $f''(x) = -\cos x$ Evaluate the terms:

• $f\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) - 1 = \frac{\sqrt{3}}{2} - 1$
• $f''\left(\frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$
• $f'\left(-\frac{\pi}{6}\right) = -\sin\left(-\frac{\pi}{6}\right) = - \left(-\sin\left(\frac{\pi}{6}\right)\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

Substitute these values into the expression:

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

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

$= 6 - 1 = 5$

The value of the expression is 5.