Question

Find the value of $\frac{\cos ({45}^0-A)}{\sin ({45}^0+A)}$

Answer 1

$\cos ({45}^0-A)=\sin [{90}^0-({45}^0-A)]$
$\because \cos \theta =\sin ({90}^0-\theta )$
$=\sin [{45}^0+A]$
$\therefore \frac{\cos ({45}^0-A)}{\sin ({45}^0+A)}=\frac{\sin ({45}^0+A)}{\sin ({45}^0+A)}=1$