Question

If A,B,C are angles of a triangle,show that:-sinBcos(C+A)+cosBsin(C+A)=0​

Answer 1

Step-by-step explanation:

Step-by-step explanation:as A, B ,C are angles of a triangle so

Step-by-step explanation:as A, B ,C are angles of a triangle so A+B+C = 180

Step-by-step explanation:as A, B ,C are angles of a triangle so A+B+C = 180A+C= 180-B ____ eq1

Step-by-step explanation:as A, B ,C are angles of a triangle so A+B+C = 180A+C= 180-B ____ eq1put values from this eq1 into question

we get sinBcos(180-B)+cosBsin(180-B)

cos(180-A)= -cosa ,, sin(180-A) =sinA

so we can say

so we can saysinB(-cosB) + cosBsinB

so we can saysinB(-cosB) + cosBsinB-sinBcosB+sinBcosB

so we can saysinB(-cosB) + cosBsinB-sinBcosB+sinBcosB=0

so we can saysinB(-cosB) + cosBsinB-sinBcosB+sinBcosB=0hence proved

Answer 2

Step-by-step explanation:

Here is your answer hope it helps