(i) LHS = (cos 20° + cos 100°) + cos 140°
= 2 cos (\(\frac{20°+100°}{2}\)) + cos (\(\frac{20°-100°}{2}\))140°
[∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))]
= 2 cos 60° cos(-40°) + cos 140°
= 2 x \(\frac{1}{2}\) x cos(-40°) + cos(180° – 140°)
[∵ cos(-θ) = cos θ, cos 60° = \(\frac{1}{2}\)
= cos 40° – cos 40°
= 0
Hence Proved.
(ii) LHS = (sin 50° – sin 70°) + sin 10°
= 2 cos (\(\frac{50+70}{2}\)) sin (\(\frac{50+70}{2}\)) + sin 10°
[∵ sin C – sin D = 2 cos(\(\frac{C+D}{2}\)) sin(\(\frac{C-D}{2}\))]
= 2 cos 60° sin(-10°) + sin 10°
= 2 x \(\frac{1}{2}\)(-sin 10°) + sin 10° [∵ sin(-θ) = -sin θ]
= -sin 10° + sin 10°
= 0
= RHS
