Question

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

Answer 1

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

we know $\sin \left(\frac{\pi }{2}\right)=1$

$\therefore A+B=\frac{\pi }{2}$

$\cos (A-B)=1$ Given

we know $\cos \left(0\right)=1$

$\therefore A-B=0$

$A=B$

$A+A=\frac{\pi }{2}$

$\Rightarrow A=\frac{\pi }{4}=B$.