Question

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

Answer 1

Given, $\sin (A+B)=1$

$\cos (A-B)=1$

$0^{\circ }<A+B\le {90}^{\circ }$ and $A\ge B$,

We know that, $\sin {90}^{\circ }=1$ and $\cos 0^{\circ }=1$

Consider,

$\sin (A+B)=1$ and $\sin {90}^{\circ }=1$

$⟹(A+B)={90}^{\circ }$ ---------------(i)

Consider,

$\cos (A-B)=1$ and $\cos 0^{\circ }=1$

$⟹(A-B)=0^{\circ }\Rightarrow A=B$ ---------------(ii)

sub (ii) in (i)

$A+A={90}^{\circ }\Rightarrow 2A={90}^{\circ }\Rightarrow A={45}^{\circ }$

$\Rightarrow A=B={45}^{\circ }$

$\therefore A=B={45}^{\circ }$