Question

If $\frac{\cos(A+B)}{\cos(A-B)} = \frac{\sin(C+D)}{\sin(C-D)}$, then $\tan A \tan B \tan C =$

• A. $0$
• B. $\tan D$
• C. $\cot D$
• D. $-\tan D$

Hint:

Whenever you see a fraction with $(A+B)$ in the numerator and $(A-B)$ in the denominator, applying Componendo and Dividendo instantly breaks the terms down into simple products of individual angles. It is a massive time-saver!

Answer 1

The Correct Option is D

Step 1: Apply componendo and dividendo.
Starting from $\dfrac{\cos(A+B)}{\cos(A-B)} = \dfrac{\sin(C+D)}{\sin(C-D)}$, add and subtract on each side.

Step 2: Use the sum to product forms.
The left becomes $\dfrac{2\cos A\cos B}{-2\sin A\sin B}$ and the right becomes $\dfrac{2\sin C\cos D}{2\cos C\sin D}$.

Step 3: Simplify.
$$-\cot A\cot B = \tan C\cot D \Rightarrow -\frac{1}{\tan A\tan B} = \frac{\tan C}{\tan D}$$
Cross multiply: $\tan A\tan B\tan C = -\tan D$.
\[ \boxed{-\tan D} \]