If sin (A + B) = 1 and cos (A – B) = 1, 0° < A + B ≤90°, A ≥ B

1.	If sin (A + B) = 1 and cos (A – B) = 1, 0° < A + B ≤90°, A ≥ B find A and B.
Answer» Given: sin (A + B) = 1 and cos (A – B) = 1 sin(A+B) = 1 Also, we know, sin 90° = 1 ⇒ sin(A+B) = sin 90° or (A+B) = 90° .............(1) Now, cos (A – B) = 1 And, we know, cos 0° = 1 ⇒ (A - B) = 0° .............(2) On solving both equations (1) and (2) , we get 2A = 90° or A = 90°/2 or A = 45° Similarly, B = 45°

If sin (A + B) = 1 and cos (A – B) = 1, 0° < A + B ≤90°, A ≥ B find A and B.

Answer»

Given: sin (A + B) = 1 and cos (A – B) = 1

sin(A+B) = 1

Also, we know, sin 90° = 1

⇒ sin(A+B) = sin 90°

or (A+B) = 90° .............(1)

Now, cos (A – B) = 1

And, we know, cos 0° = 1

⇒ (A - B) = 0° .............(2)

On solving both equations (1) and (2) ,

we get

2A = 90°

or A = 90°/2

or A = 45°

Similarly, B = 45°