Prove that: \[ \frac{\cos A + \cos B}{\cos B – \cos A} = \cot\frac{A+B}{2}\cot\frac{A-B}{2} \]
Solution
L.H.S.
\[ = \frac{\cos A + \cos B}{\cos B – \cos A} \]Use identity:
\[ \cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ \cos B – \cos A = 2\sin\frac{A+B}{2}\sin\frac{A-B}{2} \]Substitute these identities:
\[ = \frac{ 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} }{ 2\sin\frac{A+B}{2}\sin\frac{A-B}{2} } \]Cancel common factor 2:
\[ = \frac{ \cos\frac{A+B}{2} }{ \sin\frac{A+B}{2} } \cdot \frac{ \cos\frac{A-B}{2} }{ \sin\frac{A-B}{2} } \]Use identity:
\[ \cot\theta=\frac{\cos\theta}{\sin\theta} \]
\[
=
\cot\frac{A+B}{2}\cot\frac{A-B}{2}
\]
Hence Proved.