Question

sin[a+b]cos[a-b]cos c+cos d=?

Answer 1

Answer:

sin[b]-bc+d

Step-by-step explanation:

sin[a+b]cos[a-b]cos c+cos d

cos[90-(a+b) ]cos[a-b]cos c+cos d

cos[90-a-b]cos[a-b]cos c+cos d

cos[90-b]cos[-b]cos c+cos d

cos [90-b]×[-b] ×c+d

cos[-b90+b²]×c+d

cos[-bc90+b²c]+d

cos[90-b]-bc+d

sin[b]-bc+d is the answer