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Question
If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), prove that tan A tan B tan C + tan D = 0.
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Solution
cos (A + B) sin (C − D) = cos (A − B) sin (C + D)
\[\Rightarrow\][cosA cosB − sinA sinB] [sinC cosD − cosC sinD] = [cosA cosB + sinA sinB] [sinC cosD + cosC sinD]
\[\text{ Dividing both sides by }\cos A \cos B \cos C \cos D, \]
\[\frac{\left[ \cos A \cos B - \sin A \sin B \right]\left[ \sin C \cos D - \cos C \sin D \right]}{\cos A \cos B \cos C \cos D} = \frac{\left[ \cos A \cos B + \sin A \sin B \right]\left[ \sin C \cos D + \cos C \sin D \right]}{\cos A \cos B \cos C \cos D}\]
\[ \Rightarrow \frac{\left[ \cos A \cos B - \sin A \sin B \right]}{\cos A \cos B} \times \frac{\left[ \sin C\cos D - \cos C \sin D \right]}{\cos C \cos D} = \frac{\left[ \cos A \cos B + \sin A \sin B \right]}{\cos A \cos B} \times \frac{\left[ \sin C \cos D + \cos C \sin D \right]}{\cos C \cos D}\]
\[ \Rightarrow \left[ 1 - \tan A \tan B \right]\left[ \tan C - \tan D \right] = \left[ 1 + \tan A \tan B \right]\left[ \tan C + \tan D \right]\]
\[ \Rightarrow \tan C - \tan D - \tan A \tan B \tan C + \tan A \tan B \tan D = \tan C + \tan D + \tan A \tan B \tan C + \tan A \tan B \tan D\]
\[ \Rightarrow - \tan D - \tan D = \tan A \tan B \tan C + \tan A \tan B \tan C\]
\[ \Rightarrow - 2\tan D = 2\tan A \tan B \tan C\]
\[ \Rightarrow \tan A \tan B \tan C = - \tan D\]
\[ \Rightarrow \tan A \tan B \tan C + \tan D = 0\]
Hence proved.
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