Prove that: \[ \frac{ \sin(\theta+\Phi)-2\sin\theta+\sin(\theta-\Phi) }{ \cos(\theta+\Phi)-2\cos\theta+\cos(\theta-\Phi) } = \tan\theta \]
Solution
L.H.S.
\[ = \frac{ \sin(\theta+\Phi)-2\sin\theta+\sin(\theta-\Phi) }{ \cos(\theta+\Phi)-2\cos\theta+\cos(\theta-\Phi) } \]Use identities:
\[ \sin C+\sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2} \] \[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \]
\[
=
\frac{
2\sin\theta\cos\Phi-2\sin\theta
}{
2\cos\theta\cos\Phi-2\cos\theta
}
\]
Take common factor:
\[ = \frac{ 2\sin\theta(\cos\Phi-1) }{ 2\cos\theta(\cos\Phi-1) } \]Cancel common factors:
\[ = \frac{\sin\theta}{\cos\theta} \]Use identity:
\[ \tan\theta=\frac{\sin\theta}{\cos\theta} \]
\[
=
\tan\theta
\]
Hence Proved.