Question

Prove that:

\[\frac{\sin \left( \theta + \phi \right) - 2 \sin \theta + \sin \left( \theta - \phi \right)}{\cos \left( \theta + \phi \right) - 2 \cos \theta + \cos \left( \theta - \phi \right)} = \tan \theta\]

Answer 1

Consider LHS: 
\[ \frac{\sin\left( \theta + \phi \right) - 2\sin\theta + \sin\left( \theta - \phi \right)}{\cos\left( \theta + \phi \right) - 2\cos\theta + \cos\left( \theta - \phi \right)}\]
\[ = \frac{\sin\left( \theta + \phi \right) + \sin\left( \theta - \phi \right) - 2\sin\theta}{\cos\left( \theta + \phi \right) + \cos\left( \theta - \phi \right) - 2\cos\theta}\]
\[ = \frac{2\sin\left( \frac{\theta + \phi + \theta - \phi}{2} \right)\cos\left( \frac{\theta + \phi - \theta + \phi}{2} \right) - 2\sin\theta}{2\cos\left( \frac{\theta + \phi + \theta - \phi}{2} \right)\cos\left( \frac{\theta + \phi - \theta + \phi}{2} \right) - 2\cos\theta} \]
\[ = \frac{2\sin\theta\cos\phi - 2\sin\theta}{2\cos\theta\cos\phi - 2\cos\theta}\]
\[ = \frac{2\sin\theta\left[ \cos\phi - 1 \right]}{2\cos\theta\left[ \cos\phi - 1 \right]}\]
\[ = \tan\theta\]
 = RHS
Hence, RHS = LHS.