Question

$ta{n}^2\frac{\pi }{16}+ta{n}^2\frac{2\pi }{16}+....+ta{n}^2\frac{6\pi }{16}+ta{n}^2\frac{7\pi }{16}=35$

Answer 1

$(\frac{\pi }{16}+\frac{7\pi }{16})=(\frac{2\pi }{16}+\frac{6\pi }{16})=(\frac{3\pi }{16}+\frac{5\pi }{16})=\frac{\pi }{2}$
Thus these are complementary.
and $\frac{4\pi }{16}=\frac{\pi }{4}.$ If we take $\theta =\frac{\pi }{16}$
L.H.S. = $(ta{n}^2\theta +co{t}^2\theta )+(ta{n}^{2}2\theta +co{t}^{2}2\theta )+(ta{n}^{2}3\theta +co{t}^{2}3\theta )+1$
Now 1st bracket = (tan $\theta$ + cot $\theta {)}^2$ -2
$={\left(\frac{1}{sin,\theta cos\theta }\right)}^2-2=\left(\frac{2}{sin2\theta }\right)-2$
$=\frac{8}{1-cos4\theta }-2$ where $4\theta =\frac{4\pi }{16}=\frac{\pi }{4}$
$=\frac{8\sqrt{2}}{(\sqrt{2}-1)}-2=8\sqrt{2}(\sqrt{2},+1)-2.$
2nd bracket = $\frac{8}{1-cos8\theta }$ -2 = 8 - 2 = 6
3rd bracket = $\frac{8}{1-cos12\theta }=8\sqrt{2}(\sqrt{2}-1)-2$
and $ta{n}^2\frac{\pi }{4}$ = 1
$\therefore$ S = 14 + 6 + 14 + 1 = 35.