Prove that, cos10° cos20° +sin45° cos145° +sin55° cos245°=0
Solution:
L.H.S =cos10° cos20° +sin45° cos145° +sin55° cos245°
= $\frac{1}{2}$ ( 2cos10° cos20° +2sin45° cos145° +2sin55° cos245°)
=$\frac{1}{2}$ {cos(10°+20°) +cos(10°-20°)+ sin(45°+145°) +sin(45°-145°)+ sin(55°+245°) +sin(55°-245°)}
= $\frac{1}{2}$ {cos30° + cos(-10°)+ sin190°+ sin(-100°)+ sin 300° -sin 190°
= $\frac{1}{2}$ {cos30° +cos10°+ sin(180°+10°) –sin100° + sin(360°-60°)-sin(180°+10°)}
=$\frac{1}{2}$ (cos30° + cos10° –sin10° –sin100° –sin60° +sin10°)
= $\frac{1}{2}$ { cos(90°-60°) + cos10° –sin10° –sin(90°+10°) –sin60° +sin10°}
= $\frac{1}{2}$ (sin60° + cos10° –sin10° -cos10° –sin60° +sin10°)
= $\frac{1}{2}$ × 0
= 0 = R.H.S [Proved]