prove

cos20 cos100+cos100 cos140-cos140 cos200=-3/4

cos20.cos100 = $\frac{1}{2}\left(2\cos 20\cos 100\right) = \frac{1}{2}(\cos 120+\cos 80)$  ( By 2cosa.cosb = cos(a + b)cos(a - b))
Similarly,
 cos100cos140 = (1/2)(cos240 + cos40)

-cos140cos200 = -(1/2)(cos340 + cos60)

So, cos20cos100+cos100cos140-cos140cos200
= (1/2)[cos120 + cos80 + cos240 + cos40 – cos340 – cos60]

=  (1/2)[-1/2 + cos80 – 1/2 + cos40 – cos340 – 1/2]

= (1/2)[-3/2 + cos80 + cos40 – cos340]

= (1/2)[-3/2 + 2cos60cos20 – cos340]  (using cos c+cosd= 2cos(c+d)/2 cos(c-d)/2)

= (1/2)[-3/2 + cos20 – cos340]  ( as cos 60=1/2)
= (1/2)[-3/2 + cos20 – cos(360-20)] 

=  (1/2)[-3/2 + cos20 – cos(-20)]

= (1/2)[-3/2+0]
= -3/4
= RHS