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]

$Dividing both sides by \cos A \cos B \cos C \cos D, \frac{[\cos A \cos B - \sin A \sin B] [\sin C \cos D - \cos C \sin D]}{\cos A \cos B \cos C \cos D} = \frac{[\cos A \cos B + \sin A \sin B] [\sin C \cos D + \cos C \sin D]}{\cos A \cos B \cos C \cos D} \Rightarrow \frac{[\cos A \cos B - \sin A \sin B]}{\cos A \cos B} \times \frac{[\sin C \cos D - \cos C \sin D]}{\cos C \cos D} = \frac{[\cos A \cos B + \sin A \sin B]}{\cos A \cos B} \times \frac{[\sin C \cos D + \cos C \sin D]}{\cos C \cos D} \Rightarrow [1 - \tan A \tan B] [\tan C - \tan D] = [1 + \tan A \tan B] [\tan C + \tan D] \Rightarrow \tan C - \tan D - \tan A \tan B \tan C + \tan A \tan B \tan D = \tan C + \tan D + \tan A \tan B \tan C + \tan A \tan B \tan D \Rightarrow - \tan D - \tan D = \tan A \tan B \tan C + \tan A \tan B \tan C \Rightarrow - 2 \tan D = 2 \tan A \tan B \tan C \Rightarrow \tan A \tan B \tan C = - \tan D \Rightarrow \tan A \tan B \tan C + \tan D = 0 Hence proved .$

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If $c o s (A + B) s i n (C - D) = c o s (A - B) s i n (C + D), p r o v e t h a t t a n A t a n B t a n C + t a n D = 0$

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