Question

sin(A+B)/sinAcosB=cotA+tanB+1​

Answer 1

Answer:

this is the correct answer for this question

Answer 2

Step-by-step explanation:

sin(A+B)/sinAcosB=cotA+tanB+1

cos(A+B) = cosAcosB - sinAsinB

sinAsinB - cosAcosB + 1 = 0

=> -(cosAcosB - sinAsinB) + 1 = 0

=> -cos(A+B) = -1

=> cos(A+B) = 1

=> A+B = 0°, 360°...............of the form n*360°(n is an integer)

=> A = 2nπ - B

=> tanA = tan(2nπ - B)

=> tanA = -tanB

=> tanAcotA = -cotAtanB

=> -cotAtanB = 1

=> 1 + cotAtanB = 0