Question

$if \: \sin(a + b) = 1 \\ and \: \cos(a - b) = 1 \\ 0 \leqslant (a + b) \leqslant 90 \: degree \\ \: and \: a > b \: \: then \: find \\ \: value \: of \: a \: and \: b$

Answer 1

sin(A+B)=1 & cos(A-B) = 1

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

sin(A+B) = cos[90-(A+B)] = cos(A-B)

thus, 90-A-B = A-B

2A =90°

A= 45°

B = 45°

@skb

Answer 2

$\huge \textbf{hello}$

$\huge \textbf{solutions}$

$\textbf{given}$

sin(A+B) = 1

sin(A+B) = sin90.

[ Because Sin90 = 1]

A+B = 90 [equation 1]

cos(A-B) = 1

cos(A-B) = cos0

[Because Cos0= 1]

A-B = 0 [equation 2]

Add equation 1 and 2

A+B = 90  ,  A-B = 0

$\underline \textbf{now}$

=> A+B+A-B =90+0

=> A+A = 90

=> 2A = 90

=> A = 45°

$\huge \: {hence}$

$\boxed{a = 45} \: and \: \boxed{b = 45}$

$\textbf{i \: hope \: helps \: u}$

$\huge \textbf{thank \: u}$