Trigo evaluation

Express 1/(4 - $sec^{2}$ x) as cos^{2} x/(4 $cos^{2}$ x - 1) and then $cos^{2}$ x/(3 - 4 $sin^{2}$ x ) by substituting $cos^{2}$ x = 1 - $sin^{2}$ x. Then we get $cos^{2}$ x sin x/(3 sin x - 4 $sin^{3}$ x) = $cos^{2}$ x sin x/ sin 3x. By applying this to all 3 terms in the equation we get $cos^{2}$ (2pi/7) sin (2pi/7)/sin (6pi/7) + $cos^{2}$ (4pi/7) sin (4pi/7)/sin (12pi/7) + $cos^{2}$ (6pi/7) sin (6pi/7)/sin (18pi/7).

Now we can rewrite sin (6pi/7) as sin (pi - 6pi/7) = sin (pi/7), sin (12pi/7) as -sin (2pi - 12pi/7) = -sin (2pi/7) and sin (18pi/7) as sin (3pi-18pi/7) = sin (3pi/7). Now for each term we get a form $cos^{2}$ 2x sin 2x/sin x which reduces to 2 $cos^{2}$ 2x cos x as sin 2x = 2 sin x cos x. Hence we have 2 $cos^{2}$ (2pi/7) cos (pi/7) - 2 $cos^{2}$ (4pi/7) cos (2pi/7) + 2 $cos^{2}$ (6pi/7) cos (3pi/7) which gives us 2 $cos^{2}$ (2pi/7) cos (pi/7) - 2 $cos^{2}$ (3pi/7) cos (2pi/7) + 2 $cos^{2}$ (pi/7) cos (3pi/7) by invoking cos x = -cos (pi-x).

Using angle sum and difference formula we get cos (2pi/7) (cos 3pi/7 + cos pi/7) - cos (3pi/7) (cos 5pi/7 + cos pi/7) + cos (pi/7) (cos 4pi/7 + cos 2pi/7) = cos (2pi/7) (cos 3pi/7 + cos pi/7) - cos (3pi/7) (-cos 2pi/7 + cos pi/7) + cos (pi/7) (-cos 3pi/7 + cos 2pi/7) = 2 (cos pi/7 cos 2pi/7 - cos pi/7 cos 3pi/7 + cos 2pi/7 cos 3pi/7) Applying the same formula again, we get cos (3pi/7) + cos (pi/7) - cos (4pi/7) - cos (2pi/7) + cos (5pi/7) + cos (pi/7) = 2(cos pi/7 - cos 2pi/7 + cos 3pi/7) by using the good old formula cos x = -cos (pi-x) again.

Finally, we can express 2(cos pi/7 - cos 2pi/7 + cos 3pi/7) as 2(2 cos 2pi/7 cos pi/7 - cos 2pi/7) = 2cos (2pi/7)(2 cos pi/7 - 1) = 2cos (2pi/7)(-2 cos 6pi/7 - 1) = 2cos (2pi/7)(4 sin^{2} 3pi/7 - 3) using the double angle formula. Now invoking the triple angle formula, we get -2 cos (2pi/7) sin (9pi/7)/sin (3pi/7) = 2cos (2pi/7) sin (2pi/7)/sin (3pi/7) (since sin x = -sin(pi+x)) = sin (4pi/7)/sin (3pi/7) = 1 since sin x = sin (pi-x).