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Question Number 20639 by oyshi last updated on 30/Aug/17

i f \cos (A + B) \sin (C + D) = \cos (A - B)

\sin (C - D)

s o p r o o f \cot A \cot B \cot C = \cot D

Answered by Tinkutara last updated on 30/Aug/17

\frac{\sin (C + D)}{\sin (C - D)} = \frac{\cos (A - B)}{\cos (A + B)}

\frac{2 \sin C \cos D}{2 \cos C \sin D} = \frac{2 \cos A \cos B}{2 \sin A \sin B}

\tan C \cot D = \cot A \cot B

\cot A \cot B \cot C = \cot D

Answered by ajfour last updated on 30/Aug/17

\frac{\cos (A + B)}{\cos (A - B)} = \frac{\sin (C - D)}{\sin (C + D)}

\Rightarrow \frac{1 - \tan A \tan B}{1 + \tan A \tan B} = \frac{\tan C - \tan D}{\tan C + \tan D}

\Rightarrow \frac{- 2 \tan A \tan B}{2} = \frac{- 2 \tan D}{2 \tan C}

\Rightarrow \frac{1}{\tan D} = \frac{1}{\tan A \tan B \tan C} .

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