Question

$\frac{sin(\Theta +\varphi )-2sin\Theta +sin(\Theta -\varphi )}{cos(\Theta +\varphi )-2cos\Theta +cos(\theta -\varphi )}=tan\Theta$

Answer 1

$\frac{\sin (\theta +\varphi )-2\sin \theta +\sin (\theta -\varphi )}{\cos (\theta +\varphi )-2\cos \theta +\cos (\theta -\varphi )}=\tan \theta$

Using identity:

$\sin (A\pm B)=\sin A\cos B\pm \sin B\cos A$

$\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B$

Now,

Taking L.H.S.-

$\frac{\sin (\theta +\varphi )-2\sin \theta +\sin (\theta -\varphi )}{\cos (\theta +\varphi )-2\cos \theta +\cos (\theta -\varphi )}$

$=\frac{\sin \theta \cos \varphi +\cos \theta \sin \varphi -2\sin \theta +\sin \theta \cos \varphi -\cos \theta \sin \varphi }{\cos \theta \cos \varphi -\sin \theta \sin \varphi -2\cos \theta +\cos \theta \cos \varphi +\sin \theta \sin \varphi }$

$=\frac{2\sin \theta \cos \varphi -2\sin \theta }{2\cos \theta \cos \varphi -2\cos \theta }$

$=\frac{2\sin \theta (\cos \varphi -1)}{2\cos \theta (\cos \varphi -1)}$

$=\frac{\sin \theta }{\cos \theta }$

$=\tan \theta$

$=$ R.H.S.

Hence proved.