Question

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

[Proper solution needed]

Answer 1

sin (A + B) = 1
sin (A + B) = sin 90 (1)

cos (A - B) = 1
cos (A - B) = cos 0 (2)

From (1) & (2)

A + B = 90 (3)
A - B = 0 (4)

(3) + (4)
2A = 90
A= 45

Substituting this in (4) we get
B= 45

Answer 2

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

Solution :

⇒ sin (A + B) = 1 = sin = 90°

∴ A + B = 90° ..............(i)

⇒ Again, cos (A - B) = 1 = cos 0°

∴ A - B = 0° ...............(ii)

⇒ Adding (i) and (ii), we get

⇒ 2A =90°

⇒ A = $\frac{90}{2}$ = 45

∴ A = 45°

⇒ From (ii), we get

B = A = 45°

Hence, A = 45° and B = 45°