Let $$f$$ be twice differentiable function such that $$f(x) = \displaystyle\int_0^x \tan(t - x)\,dt - \int_0^x f(t)\tan t\,dt$$, $$x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$. Then $$f''\!\left(\frac{\pi}{6}\right) + 12f'\!\left(-\frac{\pi}{6}\right) + f\!\left(\frac{\pi}{6}\right)$$ is equal to :

We are given
$$f(x)=\int_{0}^{x}\tan(t-x)\,dt-\int_{0}^{x}f(t)\tan t\,dt,\qquad x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$

Step 1: Evaluate the first integral.
Put $$s=t-x.$$ When $$t=0,\;s=-x$$ and when $$t=x,\;s=0.$$ Thus $$\int_{0}^{x}\tan(t-x)\,dt=\int_{-x}^{0}\tan s\,ds.$$ Since $$\int\tan s\,ds=-\ln|\cos s|,$$ we get $$\int_{-x}^{0}\tan s\,ds=\Big[-\ln|\cos s|\Big]_{-x}^{0}=0+\ln\cos x=\ln\cos x.$$

Therefore $$f(x)=\ln\cos x-\int_{0}^{x}f(t)\tan t\,dt.\qquad -(1)$$

Step 2: Differentiate once.
Differentiate both sides of $$(1)$$ with respect to $$x$$ (using Leibniz rule for the integral): $$f'(x)=\frac{d}{dx}\bigl(\ln\cos x\bigr)-f(x)\tan x.$$ Because $$\dfrac{d}{dx}\ln\cos x=-\tan x,$$ we get $$f'(x)=-\tan x-f(x)\tan x=-\tan x\bigl(1+f(x)\bigr).\qquad -(2)$$

Step 3: Differentiate again.
Differentiate $$(2):$$ $$f''(x)=-\sec^{2}x\bigl(1+f(x)\bigr)-\tan x\,f'(x).$$ Substitute $$f'(x)$$ from $$(2):$$ $$f''(x)=-\sec^{2}x\bigl(1+f(x)\bigr)-\tan x\bigl[-\tan x\,(1+f(x))\bigr].$$ Since $$\sec^{2}x=\tan^{2}x+1,$$ the bracket simplifies:
$$-\sec^{2}x+\tan^{2}x=-1.$$ Hence $$f''(x)=-(1+f(x))\quad\Longrightarrow\quad f''(x)+f(x)+1=0.\qquad -(3)$$

Step 4: Initial conditions.
From $$(1)$$ at $$x=0$$: $$f(0)=\ln\cos0=0.$$ From $$(2)$$ at $$x=0$$: $$f'(0)=-\tan0\,(1+f(0))=0.$$

Step 5: Solve the differential equation.
The homogeneous part of $$(3)$$ is $$f_h''+f_h=0,$$ giving $$f_h=C_1\cos x+C_2\sin x.$$ Seek a particular solution $$f_p=A$$ (constant). Substituting in $$(3)$$: $$0+A+1=0\;\Rightarrow\;A=-1.$$ Thus $$f(x)=C_1\cos x+C_2\sin x-1.$$ Apply the initial conditions: $$f(0)=C_1-1=0\;\Longrightarrow\;C_1=1,$$ $$f'(x)=-C_1\sin x+C_2\cos x,$$ so $$f'(0)=C_2=0.$$ Therefore $$f(x)=\cos x-1,\qquad f'(x)=-\sin x,\qquad f''(x)=-\cos x.$$

Step 6: Compute the required expression.
At $$x=\frac{\pi}{6}:$$ $$f\!\left(\frac{\pi}{6}\right)=\cos\frac{\pi}{6}-1=\frac{\sqrt3}{2}-1,$$ $$f''\!\left(\frac{\pi}{6}\right)=-\cos\frac{\pi}{6}=-\frac{\sqrt3}{2}.$$ At $$x=-\frac{\pi}{6}:$$ $$f'\!\left(-\frac{\pi}{6}\right)=-\sin\!\left(-\frac{\pi}{6}\right)=\frac12.$$ Hence $$f''\!\left(\frac{\pi}{6}\right)+12\,f'\!\left(-\frac{\pi}{6}\right)+f\!\left(\frac{\pi}{6}\right)$$ $$=\left(-\frac{\sqrt3}{2}\right)+12\!\left(\frac12\right)+\left(\frac{\sqrt3}{2}-1\right)= -\frac{\sqrt3}{2}+6+\frac{\sqrt3}{2}-1=5.$$

Therefore the required value is 5.