tan ^2 frac{pi}{16}+tan ^2 frac{2 pi}{16}+tan | vedclass

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(A) Let $K = \sum_{k=1}^{7} 	an^2 \frac{k\pi}{16}$.Using the property $	an^2 	heta + \cot^2 	heta = (	an 	heta + \cot 	heta)^2 - 2 = \frac{4}{\

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(A) Let $K = \sum_{k=1}^{7} 	an^2 \frac{k\pi}{16}$.Using the property $	an^2 	heta + \cot^2 	heta = (	an 	heta + \cot 	heta)^2 - 2 = \frac{4}{\sin^2 2	heta} - 2 = \frac{8}{1 - \cos 4	heta} - 2$.We group the terms as:$K = (	an^2 \frac{\pi}{16} + 	an^2 \frac{7\pi}{16}) + (	an^2 \frac{2\pi}{16} + 	an^2 \frac{6\pi}{16}) + (	an^2 \frac{3\pi}{16} + 	an^2 \frac{5\pi}{16}) + 	an^2 \frac{4\pi}{16}$.Since $	an \frac{7\pi}{16} = \cot \frac{\pi}{16}$,etc.,this becomes:$K = (	an^2 \frac{\pi}{16} + \cot^2 \frac{\pi}{16}) + (	an^2 \frac{\pi}{8} + \cot^2 \frac{\pi}{8}) + (	an^2 \frac{3\pi}{16} + \cot^2 \frac{3\pi}{16}) + 1$.Using the identity $	an^2 	heta + \cot^2 	heta = \frac{8}{1 - \cos 4	heta} - 2$:For $	heta = \frac{\pi}{16}$,$4	heta = \frac{\pi}{4}$,sum $= \frac{8}{1 - 1/\sqrt{2}} - 2 = 8(2 + \sqrt{2}) - 2 = 14 + 8\sqrt{2}$.For $	heta = \frac{\pi}{8}$,$4	heta = \frac{\pi}{2}$,sum $= \frac{8}{1 - 0} - 2 = 6$.For $	heta = \frac{3\pi}{16}$,$4	heta = \frac{3\pi}{4}$,sum $= \frac{8}{1 + 1/\sqrt{2}} - 2 = 8(2 - \sqrt{2}) - 2 = 14 - 8\sqrt{2}$.Summing these: $K = (14 + 8\sqrt{2}) + 6 + (14 - 8\sqrt{2}) + 1 = 35$.

$\tan ^2 \frac{\pi}{16}+\tan ^2 \frac{2 \pi}{16}+\tan ^2 \frac{3 \pi}{16}+\tan ^2 \frac{4 \pi}{16}+\tan ^2 \frac{5 \pi}{16}+\tan ^2 \frac{6 \pi}{16}+\tan ^2 \frac{7 \pi}{16} = ?$

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