Question

\[\text{ If }\frac{\cos (A - B)}{\cos (A + B)} + \frac{\cos (C + D)}{\cos (C - D)} = 0, \text {Prove that }\tan A \tan B \tan C \tan D = - 1\]

 

Answer 1

We have, 
\[\frac{\cos \left( A - B \right)}{\cos \left( A + B \right)} + \frac{\cos \left( C + D \right)}{\cos \left( C - D \right)} = 0\]
\[ \Rightarrow \frac{\cos \left( A - B \right) \cos \left( C - D \right) + \cos \left( C + D \right) \cos \left( A + B \right)}{\cos \left( A + B \right) \cos \left( C - D \right)} = 0\]
\[ \Rightarrow \cos \left( A - B \right) \cos \left( C - D \right) + \cos \left( C + D \right) \cos \left( A + B \right) = 0\]
\[ \Rightarrow \cos \left( A - B \right) \cos \left( C - D \right) = - \cos \left( C + D \right) \cos \left( A + B \right)\]
\[ \Rightarrow \left[ \cos A \cos B + \sin A \sin B \right]\left[ \cos C \cos D + \sin C \sin D \right] = - \left[ \cos C \cos D - \sin C \sin D \right]\left[ \cos A \cos B - \sin A \sin B \right]\]
\[\text{ Dividing both sides by }\cos A \cos B \cos C \cos D \text{ we get, }\]
\[\frac{\left[ \cos A \cos B + \sin A\sin B \right]\left[ \cos C\cos D + \sin C\sin D \right]}{\cos A\cos B\cos C\cos D} = - \frac{\left[ \cos C \cos D - \sin C\sin D \right]\left[ \cos A\cos B - \sin A\sin B \right]}{\cos A\cos B\cos C\cos D}\]
\[ \Rightarrow \frac{\left[ \cos A \cos B + \sin A\sin B \right]}{\cos A\cos B} \times \frac{\left[ \cos C\cos D + \sin C\sin D \right]}{\cos C\cos D} = - \frac{\left[ \cos C \cos D - \sin C\sin D \right]}{\cos C\cos D} \times \frac{\left[ \sin C\cos A\cos B - \sin A\sin B \right]}{\cos A\cos B}\]
\[ \Rightarrow \left[ 1 + \tan A\tan B \right]\left[ 1 + \tan C\tan D \right] = \left[ \tan C\tan D - 1 \right]\left[ 1 - \tan A\tan B \right]\]
\[ \Rightarrow 1 + \tan C\tan D + \tan A\tan B + \tan A\tan B\tan C\tan D = \tan C\tan D - \tan A\tan B\tan C\tan D + \tan A\tan B\tan D - 1 + \tan A\tan B\]
\[ \Rightarrow 2\tan A\tan B\tan C\tan D = - 2\]
\[ \Rightarrow \tan A\tan B\tan C\tan D = - 1\]
Hence proved.