Question

If tan =1and sinB ==1/root2 find the value of cos(A +B) where A,B are acute angles​

Answer 1

Step-by-step explanation:

cos(A+B) = ??????

formula

cos(A+B) = cosAcosB - sinAsinB

Given:

sinB = 1/√2

formula

cosB = √(1-sin^2B)

cosB = √{1-(1/√2)^2}

cosB = √(1-1/2)

cosB = 1/√2

tanA = 1

now

tanA = sinA/cosA

sinA/cosA = 1

sinA = cosA

cos(A+B) = cosAcosB - sinAsinB

cos(A+B) = sinA×1/√2 - sinA×1/√2

cos(A+B) = sinA/√2 - sinA/√2

cos(A+B) = 0

Answer 2

Step-by-step explanation:

tan A= 1

tan 45°= 1

sin B = 1/ root 2

sin 45° = 1/ root 3

cos (45°+45°) = 0