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: Understanding the Question:
We are given a ratio of trigonometric sum and difference formulas and need to find the product of $\tan A$, $\tan B$, and $\tan C$.

Step 2: Key Formula or Approach:
The most efficient way to simplify ratios of the form $\frac{x}{y} = \frac{p}{q}$ in trigonometry is by using Componendo and Dividendo: $\frac{x+y}{x-y} = \frac{p+q}{p-q}$.
Then apply the standard factor formulas (sum to product transformations):
$\cos X + \cos Y = 2 \cos\left(\frac{X+Y}{2}\right) \cos\left(\frac{X-Y}{2}\right)$
$\cos X - \cos Y = -2 \sin\left(\frac{X+Y}{2}\right) \sin\left(\frac{X-Y}{2}\right)$
$\sin X + \sin Y = 2 \sin\left(\frac{X+Y}{2}\right) \cos\left(\frac{X-Y}{2}\right)$
$\sin X - \sin Y = 2 \cos\left(\frac{X+Y}{2}\right) \sin\left(\frac{X-Y}{2}\right)$

Step 3: Detailed Explanation:
Given:
$$\frac{\cos(A+B)}{\cos(A-B)} = \frac{\sin(C+D)}{\sin(C-D)}$$ Apply Componendo and Dividendo:
$$\frac{\cos(A+B) + \cos(A-B)}{\cos(A+B) - \cos(A-B)} = \frac{\sin(C+D) + \sin(C-D)}{\sin(C+D) - \sin(C-D)}$$ Substitute the factor formulas:
$$\frac{2 \cos A \cos B}{-2 \sin A \sin B} = \frac{2 \sin C \cos D}{2 \cos C \sin D}$$ Simplify the fractions:
$$-\cot A \cot B = \tan C \cot D$$ Recall that $\cot \theta = \frac{1}{\tan \theta}$:
$$-\frac{1}{\tan A \tan B} = \frac{\tan C}{\tan D}$$ Cross-multiply to isolate the required terms:
$$-\tan D = \tan A \tan B \tan C$$

Step 4: Final Answer:
The value of $\tan A \tan B \tan C$ is $-\tan D$, matching option (D).