Pokaži dokazati niži identitet? 1 / cos290 + 1 / (sqrt3sin250) = 4 / sqrt3

# LHS = 1 / (cos290 ^ ') + 1 / (sqrt3sin250 ^') #

# = 1 / (cos (360-70) ^ ') + 1 / (sqrt3sin (180 + 70) ^') #

# = 1 / (cos70 ^ ') - 1 / (sqrt3sin70 ^') #

# = (Sqrt3sin70 ^ '- cos70 ^') / ('^ sqrt3sin70 cos70 ^') #

# = 1 / sqrt3 (2 ^ {sqrt3sin70 '- ^ cos70 @}) / (' ^ 2sin70 cos70 ^ ') #

# = 1 / sqrt3 (2 * 2 {sin70 ^ '* (sqrt3 / 2) -cos70 ^' * (1/2)}) / (sin140 ^ ') #

# = 1 / sqrt3 (4- {sin70 ^ '* cos30 ^' - ^ '* cos70 sin30 ^ @}) / (sin (180-40) ^') #

# = 1 / sqrt3 (4- {sin (70-30) ^ '}) / (sin40 ^') = 1 / sqrt3 (4- {poništavanje (sin40 ^ ')}) / otkazivanje ((sin40 ^')) = 4 / sqrt3 = RHS #

NAPOMENA #cos (360-A) ^ @ = cosA i sin (180 + A) ^ @ = - sinA #

# 1 / cos290 + 1 / (sqrt3sin250) #

# = 1 / cos (270 + 20) + 1 / (sqrt3sin (270-20)) #

# = 1 / sin20 - 1 / (sqrt3cos20) #

# = (sqrt3cos20-sin20) / (sqrt3sin20cos20) #

# = 2 / sqrt3 (sqrt3 / 2cos20-1 / 2sin20) / (sin20cos20) #

# = 4 / sqrt3 (sin60cos20-cos60sin20) / (2sin20cos20) #

# = 4 / sqrt3 sin (60-20) / (2sin20cos20) #

# = 4 / sqrt3 sin40 / sin40 #

# = 4 / sqrt3 #