Apakah kos (arctan (3)) + dosa (arctan (4)) sama?

Jawapan:

#cos (arctan (3)) + sin (arctan (4)) = 1 / sqrt (10) + 4 / sqrt (17) #

Penjelasan:

Biarkan # tan ^ -1 (3) = x #

kemudian # rarrtanx = 3 #

# rarrsecx = sqrt (1 + tan ^ 2x) = sqrt (1 + 3 ^ 2) = sqrt (10) #

# rarrcosx = 1 / sqrt (10) #

# rarrx = cos ^ (- 1) (1 / sqrt (10)) = tan ^ (- 1) (3) #

Juga, mari #tan ^ (- 1) (4) = y #

kemudian # rarrtany = 4 #

# rarrcoty = 1/4 #

# rarrcscy = sqrt (1 + cot ^ 2y) = sqrt (1+ (1/4) ^ 2) = sqrt (17) / 4 #

# rarrsiny = 4 / sqrt (17) #

# rarry = sin ^ (- 1) (4 / sqrt (17)) = tan ^ (- 1) 4 #

Sekarang, #rarrcos (tan ^ (- 1) (3)) + dosa (tan ^ (- 1) tan (4)) #

#rarrcos (cos ^ -1 (1 / sqrt (10))) + sin (sin ^ (- 1) (4 / sqrt (17))) = 1 / sqrt (10) + 4 / sqrt (17)