Question

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

• A. $\sin \alpha -\cos \alpha =\pm \sqrt{2}\sin \theta$
• B. $\sin \alpha +\cos \alpha =\pm \sqrt{2}\cos \theta$
• C. $\cos 2\theta =\sin 2\alpha$
• D. $\sin 2\theta +\cos 2\alpha =0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\cos 2\theta =\sin 2\alpha$
B $\sin \alpha +\cos \alpha =\pm \sqrt{2}\cos \theta$
C $\sin \alpha -\cos \alpha =\pm \sqrt{2}\sin \theta$
D $\sin 2\theta +\cos 2\alpha =0$

${\cos }^2\theta =\frac{1}{{sec}^2\theta }=\frac{1}{1+{\tan }^2\theta }$
Using $\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }=\tan \theta$
We get
${\cos }^2\theta =\frac{1}{1+{\left(\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }\right)}^2}=\frac{{(\sin \alpha +\cos \alpha )}^2}{2}$
$\Rightarrow \sin \alpha +\cos \alpha =\pm \sqrt{2}\cos \theta$
And
${\sin }^2\theta =1-{\cos }^2\theta =\frac{{(\sin \alpha -\cos \alpha )}^2}{2}\Rightarrow \sin \alpha -\cos \alpha =\pm \sqrt{2}\sin \theta$
Now,
$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta =\frac{{(\sin \alpha +\cos \alpha )}^2}{2}-\frac{{(\sin \alpha -\cos \alpha )}^2}{2}=\sin 2\alpha$
And
$\sin 2\theta =2\sin \theta \cos \theta =-\cos 2\alpha$