Question

Solve
$\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }$

Answer 1

$\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }$

$=\frac{\sin \alpha -\sin (\frac{\pi }{2}-\alpha )}{\sin \alpha +\sin (\frac{\pi }{2}-\alpha )}$ since $\sin (\frac{\pi }{2}-\alpha )=\cos \alpha$

Using transformation angle formula $\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$ and $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

$=\frac{2\cos \left(\frac{\alpha +\frac{\pi }{2}-\alpha }{2}\right)\sin \left(\frac{\alpha -\frac{\pi }{2}+\alpha }{2}\right)}{2\sin \left(\frac{\alpha +\frac{\pi }{2}-\alpha }{2}\right)\cos \left(\frac{\alpha -\frac{\pi }{2}+\alpha }{2}\right)}$

$=\frac{\cos \frac{\pi }{4}\sin (\alpha -\frac{\pi }{4})}{\sin \frac{\pi }{4}\cos (\alpha -\frac{\pi }{4})}$

We know that $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ and $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$

$=\frac{\sin (\alpha -\frac{\pi }{4})}{\cos (\alpha -\frac{\pi }{4})}$