Question

If $\cos (A+B)\sin (C+D)=\cos (A-B)\sin (C-D),$ then show that $\cot A\cot B\cot C=\cot D$.

Answer 1

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

rearragnging and using trigonometric identities

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

$\Rightarrow \frac{\cos A\cos B-\sin A\sin B}{\cos A\cos B+\sin A\sin B}=\frac{\sin C\cos D-\cos C\sin D}{\sin C\cos D-\cos C\sin D}$

$\Rightarrow \frac{2\cos A\cos B}{-2\sin A\sin B}=\frac{2\sin C\cos D}{-2\cos C\sin D}\Rightarrow \cot A\cot B=\tan C\tan D\Rightarrow \cot A\cot B\cot C=\cot D$