Question

If sin (A + B) = 1 and cos (A - B) = 1, $0^{\circ }\le (A+B)\le {90}^{\circ }$ and A > B then find A and B.

Answer 1

sin(A+B) = 1

sin(A+B) = sin90

A+B = 90 [equation 1]

cos(A-B) = 1

cos(A-B) = cos 0

A-B = 0 [equation 2]

add equation (1) and (2) we get

2A=90

$A={45}^o$
put the value of A in equation (2)

A-B=0

45-B=0

$B={45}^o$

$A=B={45}^o$