<s>[INST] Finding the value of a given trigonometric series. 
Find the value of $\tan^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\tan^2\dfrac{4\pi}{16}+\tan^2\dfrac{5\pi}{16}+\tan^2\dfrac{6\pi}{16}+\tan^2\dfrac{7\pi}{16}.$

My attempts: 
I converted the given series to a simpler form: 
$\tan^2\dfrac{\pi}{16}+\cot^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\cot^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\cot^2\dfrac{3\pi}{16}+1.$
Then I found the following values because I already knew the values of $\sin22.5^{\circ}$ and $\cos22.5^{\circ}$: 
$\cos^2(\frac{\pi}{16})= \dfrac{2+\sqrt{2+\sqrt2}}{4}$
$\sin^2(\frac{\pi}{16})= \dfrac{2-\sqrt{2+\sqrt2}}{4}$
$\sin^2(\frac{\pi}{8})= \dfrac{2-\sqrt2}{4}$
$\cos^2(\frac{\pi}{8})= \dfrac{2+\sqrt2}{4}$
However, at this stage I feel that my method of solving this problem is unnecessarily long and complicated. Could you guide me with a simpler approach to this question?
 [/INST] Use the following identity:
$$ \cot x - \tan x = 2 \cot 2x$$
Square this
$$ \cot^2 x + \tan^2 x = 2 + 4 \cot^2 2x$$
Use this again with your idea of grouping terms as $\tan^2 x + \cot^2x$, and you will see that all you need to know is the value of $$\cot \frac{4\pi}{16} = \cot \frac{\pi}{4}$$
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