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

$$if\cos (A+B)\sin (C+D)=\cos (A-B)$$$$\sin (C-D)$$$$soproof\cot A\cot B\cot C=\cot D$$

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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