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: Start from the given ratio.
We have $\dfrac{\cos(A+B)}{\cos(A-B)}=\dfrac{\sin(C+D)}{\sin(C-D)}$ and want $\tan A\tan B\tan C$.
Step 2: Apply componendo and dividendo.
This turns each side into a ratio of (sum) over (difference): $\dfrac{\cos(A+B)+\cos(A-B)}{\cos(A+B)-\cos(A-B)}=\dfrac{\sin(C+D)+\sin(C-D)}{\sin(C+D)-\sin(C-D)}$.
Step 3: Use sum-to-product on the left.
Top: $\cos(A+B)+\cos(A-B)=2\cos A\cos B$. Bottom: $\cos(A+B)-\cos(A-B)=-2\sin A\sin B$. So left side $=-\cot A\cot B$.
Step 4: Do the same on the right.
Top: $\sin(C+D)+\sin(C-D)=2\sin C\cos D$. Bottom: $\sin(C+D)-\sin(C-D)=2\cos C\sin D$. So right side $=\dfrac{\tan C}{\tan D}$.
Step 5: Equate the two simplified sides.
$-\dfrac{1}{\tan A\tan B}=\dfrac{\tan C}{\tan D}$.
Step 6: Cross multiply.
$-\tan D=\tan A\tan B\tan C$, which is option (4). \[ \boxed{\tan A\tan B\tan C=-\tan D} \]