Trigonometric Tables Are Insufficient!

Since $\tan^2x + \tan^2(\tfrac12\pi - x) \; = \; \tan^2x + \cot^2x \; = \; \frac{\sin^4x + \cos^4x}{\sin^2x\cos^2x} \; = \; \frac{(\sin^2x + \cos^2x)^2 - 2\sin^2x\cos^2x}{\sin^2x\cos^2x} \; = \; 4\mathrm{cosec}^22x - 2$ and $\mathrm{cosec}^2x + \mathrm{cosec}^2(\tfrac12\pi-x) \; = \; \mathrm{cosec}^2x + \sec^2x \; = \; \frac{\cos^2x + \sin^2x}{\sin^2x\cos^2x} \; = \; 4\mathrm{cosec}^22x$ the sum is equal to $4\mathrm{cosec}^2\tfrac18\pi - 2 + 4\mathrm{cosec}^2\tfrac38\pi - 2 \; = \; 16\mathrm{cosec}^2\tfrac14\pi - 4 \; = \; 16\times2 - 4 \; = \; \boxed{28}$

Relevant wiki: Chebyshev Polynomials - Definition and Properties

Let $I$ denote the value of this expression, then by applying the property that $\tan x = -\tan(\pi - x)$ , we can rewrite $I$ as

$\tan^{2} \frac{15\pi}{16} + \tan^{2} \frac{13\pi}{16} + \tan^{2} \frac{11\pi}{16} + \tan^{2} \frac{9\pi}{16}.$

Adding the original expression and the expression above gives

$2I = \tan^{2} \frac{\pi}{16} + \tan^{2} \frac{3\pi}{16} + \tan^{2} \frac{5\pi}{16} + \tan^{2} \frac{7\pi}{16} + \tan^{2} \frac{9\pi}{16} + \tan^{2} \frac{11\pi}{16} + \tan^{2} \frac{13\pi}{16} + \tan^{2} \frac{15\pi}{16}$

Now, consider the equation $\cos(8x) = 0$ . It has roots $\dfrac{\pi}{16} , \dfrac{3\pi}{16}, \dfrac{5\pi}{16}, \dfrac{7\pi}{16}, \dfrac{9\pi}{16}, \dfrac{11\pi}{16}, \dfrac{13\pi}{16}, \dfrac{15\pi}{16}$ .

By applying the double angle formula $\cos(2A) = 2\cos^2 A- 1$ repeatedly, we can rewrite $\cos(8x)$ as a polynomial of $\cos x$ ,

$\cos(8x) =2 ( 2 (2\cos^2 x - 1)^2 - 1)^2 - 1 = 128\cos^8 x - 256\cos^6 x + 160\cos^4 x - 32\cos^2 x + 1.$

Let $y = \cos x$ , then $128y^8 - 256y^6 + 160y^4 - 32y^2 + 1 = 0$ has roots $\cos \dfrac{\pi}{16} , \cos \dfrac{3\pi}{16}, \cos \dfrac{5\pi}{16}, \cos\dfrac{7\pi}{16}, \cos\dfrac{9\pi}{16}, \cos \dfrac{11\pi}{16}, \cos \dfrac{13\pi}{16}, \cos \dfrac{15\pi}{16}$ .

Likewise, let $z = \dfrac1y$ , then $128 \left(\dfrac1z\right)^8 - 256\left(\dfrac1z\right)^6 + 160\left(\dfrac1z\right)^4 - 32\left(\dfrac1z\right)^2 + 1 = 0 \quad \Leftrightarrow z^8 - 32z^6 + 160z^4 - 256z^2 + 128 = 0$ has roots

$\sec \dfrac{\pi}{16} , \sec \dfrac{3\pi}{16}, \sec \dfrac{5\pi}{16}, \sec\dfrac{7\pi}{16}, \sec\dfrac{9\pi}{16}, \sec \dfrac{11\pi}{16}, \sec \dfrac{13\pi}{16}, \sec \dfrac{15\pi}{16}$ .

By Vieta's formula,

$\begin{aligned} 0^2 - 2(-32) &=& \sec^2 \dfrac{\pi}{16} + \sec^2 \dfrac{3\pi}{16}+ \sec^2 \dfrac{5\pi}{16} + \sec^2\dfrac{7\pi}{16}+ \sec^2\dfrac{9\pi}{16}+ \sec^2 \dfrac{11\pi}{16} + \sec^2 \dfrac{13\pi}{16} + \sec^2 \dfrac{15\pi}{16} \\ 64 &=& \tan^2 \dfrac{\pi}{16} + \tan^2 \dfrac{3\pi}{16}+ \tan^2 \dfrac{5\pi}{16} + \tan^2 \dfrac{7\pi}{16}+ \tan^2 \dfrac{9\pi}{16}+ \tan^2 \dfrac{11\pi}{16} + \tan^2 \dfrac{13\pi}{16} + \tan^2 \dfrac{15\pi}{16} + 8 \\ 64 - 8 & =& 2I \\ I &=& \boxed{28} \; . \end{aligned}$

i will share a slightly shorter approach

let pi/16 =x

lets club first and last term

tan^2(x)+cot^2(x) =[ sin^4(x)+cos^4(x) ]/[sinxcosx]^2

Using sin^4(x)+cos^4(x) = 1-2(sinxcosx)^2

similarly doing for second and third term

And solving further we get

4(cosec^2(2x)+cosec^2(6x)-1)

Now use cosec^2(y)= 1+cot^2(y)

cot^2(2x) = 3+2*root2

cot^2(6x) = 3-2*root 2

$\tan^{2} \frac{\pi}{16} + \tan^{2} \frac{3\pi}{16} + \tan^{2} \frac{5\pi}{16} + \tan^{2} \frac{7\pi}{16}$ $= \tan^{2} \frac{\pi}{16} + \cot^{2}\frac{\pi}{16} + \tan^{2} \frac{3\pi}{16} + \cot^{2}\frac{3\pi}{16}$ $=\frac{\sin^2\frac{\pi}{16}}{\cos^2\frac{\pi}{16}}+ \frac{\cos^2\frac{\pi}{16}}{\sin^2\frac{\pi}{16}} +\frac{\sin^2\frac{3\pi}{16}}{\cos^2\frac{3\pi}{16}}+ \frac{\cos^2\frac{3\pi}{16}}{\sin^2\frac{3\pi}{16}}$ $=\frac{\sin^4\frac{\pi}{16} + \cos^4\frac{\pi}{16} } {\sin^2\frac{\pi}{16} \times \cos^2\frac{\pi}{16} } +\frac{\sin^4\frac{3\pi}{16} + \cos^4\frac{3\pi}{16} } {\sin^2\frac{3\pi}{16} \times \cos^2\frac{3\pi}{16} }$ $=\frac {1 - (2\sin^2\frac{\pi}{16}\times\cos^2\frac{\pi}{16})}{\sin^2\frac{\pi}{16}\times\cos^2\frac{\pi}{16}} + \frac {1 - (2\sin^2\frac{3\pi}{16}\times\cos^2\frac{3\pi}{16})}{\sin^2\frac{3\pi}{16}\times\cos^2\frac{3\pi}{16}}$ $= \frac{4}{\sin^2\frac{2\pi}{16}} + \frac{4}{\sin^2\frac{6\pi}{16}} -4$ $=4 \times \left( \frac{sin^2\frac{3\pi}{8} + sin^2\frac{\pi}{8}}{sin^2\frac{3\pi}{8}\times sin^2\frac{\pi}{8}} \right) - 4$ $=16 \times \frac{ [1- (cos^2\frac{3\pi}{8} - sin^2\frac{\pi}{8})]}{(2 \space sin\frac{3\pi}{8}\times sin\frac{\pi}{8})^2} - 4$ $=16 \times \frac{ [1- (cos\frac{\pi}{2} \times cos\frac{\pi}{4})]}{(cos\frac{\pi}{4} - cos \frac{\pi}{2})^2} - 4$ $=\frac{16}{(\frac{1}{\sqrt2})^2} - 4$ $=32 - 4$ $=\boxed{28}$ .

P.S : I have used the identities:

$(a)\cos^2x - \sin^2y = \cos(x+y) \times \cos(x-y)$

$(b)\space 2\sin x \sin y = \cos(x-y) - \cos(x+y)$

$(c)\sin^4x + \cos^4x = 1 - 2 \sin^2x \cos^2x$

$(d)\space 2 \sin x \cos x = \sin2x$

$(e) \cot (\frac{\pi}{2} - x ) = \tan x$ .

$\color{#20A900}{\cos 2A=2\cos^2 A-1=1-2\sin^2 A=\cos^2 A - \sin^2 A.\\ \therefore\ \ \ \ \cos \frac \pi 8=\sqrt{\dfrac{\cos\frac \pi 4+1} 2}=\sqrt{ \dfrac{\frac {\sqrt2} 2+1} 2 }=\frac 1 2 *\sqrt{2+\sqrt2}..........(1)\\ Also\ \ \sin \frac \pi 8=\sqrt{\dfrac{1 - \sin\frac \pi 4} 2}=\sqrt{\dfrac{1 -\frac {\sqrt2} 2 } 2}=\frac 1 2 *\sqrt{2-\sqrt2}..........(2)\\ \implies\ \ \dfrac 1 {\tan^2 \frac \pi 8}=(\dfrac {(2)}{(1)})^2 =\dfrac {2 + \sqrt2} { 2-\sqrt2` }=\dfrac{6 + 2*2*\sqrt2} 2= 3 + 2 \sqrt2 ...............\color{#3D99F6}{**} \\ Similarly\ \ \ \tan^2 \frac \pi 8=3 -2\sqrt2..........\color{#E81990}{@@} }\\ \\ \ \ \ \\ \ \ \ \ \\ \tan(\frac {7 \pi}{16})= - \dfrac 1 {\tan(\frac \pi 2 - \frac {7 \pi} {16})} = \dfrac 1 {\tan( - \frac \pi {16}) }.\\ \tan\frac {5 \pi} {16}=\dfrac 1 { \tan ( \frac \pi 2 -\frac {5\pi} {16})}=\dfrac 1 {\tan(\frac{3 \pi }{16})}.\\ And\ \ tan^2 \frac{3\pi} 8= \tan^2(-(\pi - \frac{3\pi} 8) )=\tan^2 \frac{\pi} 8.\\ \\ \ \ \ \ \\ \ \ \ \ \\ \color{#EC7300}{Also\ \ \tan^2 x + \dfrac 1{ \tan^2 x} \; = \dfrac{\sin^2 x}{\cos^2 x} +\dfrac{ \cos^2 x}{\sin^2 x} \\ =\dfrac{\sin^4 x + \cos^4 x}{\sin^2 x\cos^2 x} \; = \; \dfrac{(\cos^2 x - \sin^2 x)^2+ 2\sin^2 x*\cos^2 x}{\sin^2 x\cos^2 x}\\ =\dfrac{4( \cos^2 2x)^2 }{\sin^2 2x}+2 \; =\dfrac 4 {\tan^2 2x} +2.}\\ \\ \ \ \ \ \\ \ \ \ \\ Since \ \ 2x=\frac \pi 8,\ \ \ and\ \ \ \color{#3D99F6}{**},\ \ \ \ \color{#E81990}{@@}; \ \ \\ \tan^{2} \frac{\pi}{16} + \tan^{2} \frac{3\pi}{16} + \tan^{2} \frac{5\pi}{16} + \tan^{2} \frac{7\pi}{16} \\ = \tan^{2} \frac{\pi}{16} + \dfrac 1 { \tan^2(-\frac \pi {16}) } \ \ \ \ + \ \ \ \ \dfrac 1 {\tan^2(\frac{3 \pi }{16})} + \tan^2 \frac{3\pi}{16} \\ =\dfrac 4 {\tan^2 \frac{\pi} 8} +2 \ \ \ +\ \ \ \dfrac 4 {\tan^2 \frac{3\pi} 8} +2 \\ =4(3 + 2\sqrt2) + 2\ \ \ +\ \ \ 4(3 - 2\sqrt2) + 2= \Large\ \ \color{#D61F06}{28}.\\$

Interesting to note for angle $\dfrac \pi {2^n}$
$\color{#E81990}{Applying\ Half\ Angle\ Formula,\\ Cos\frac \pi 4=\dfrac{\sqrt2} 2.\\ Cos\frac \pi 8=\dfrac{\sqrt{2+\sqrt2} } 2.\\ Cos\frac \pi {16}=\dfrac{\sqrt{2+\sqrt{2+\sqrt2} }} 2.\\ Cos\frac \pi {32}=\dfrac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2} } } } 2. \\ Sin\frac \pi 4=\dfrac{\sqrt2} 2.\\ Sin\frac \pi 8=\dfrac{\sqrt{2 - \sqrt2} } 2.\\ Sin\frac \pi {16}=\dfrac{\sqrt{2 - \sqrt{2+\sqrt2} }} 2.\\ Sin\frac \pi {32}=\dfrac{\sqrt{2- \sqrt{2+\sqrt{2+\sqrt2} } } } 2. }\\ \text{And so on. }$