If `cos2B=(cos(A+C))/(cos(A-C))`, then tan A, tan B, tan C are in

Options

A. A.P.
B. GP.
C. H.P.
D. none of these

Answer

To solve the problem, we need to show that if \( \cos 2B = \frac{\cos(A+C)}{\cos(A-C)} \), then \( \tan A, \tan B, \tan C \) are in geometric progression (GP). Here’s a step-by-step solution:

### Step 1: Start with the given equation
We have:
\[ \cos 2B = \frac{\cos(A+C)}{\cos(A-C)} \]

### Step 2: Apply the componendo and dividendo
Using the componendo and dividendo method, we can rewrite the equation as:
\[ \frac{\cos 2B + 1}{\cos 2B - 1} = \frac{\cos(A+C) + \cos(A-C)}{\cos(A+C) - \cos(A-C)} \]

### Step 3: Expand the right side using cosine addition formulas
We know:
\[ \cos(A+C) + \cos(A-C) = 2 \cos A \cos C \]
\[ \cos(A+C) - \cos(A-C) = -2 \sin A \sin C \]
So we can rewrite the equation as:
\[ \frac{\cos 2B + 1}{\cos 2B - 1} = \frac{2 \cos A \cos C}{-2 \sin A \sin C} \]

### Step 4: Simplify the equation
This simplifies to:
\[ \frac{\cos 2B + 1}{\cos 2B - 1} = -\frac{\cos A \cos C}{\sin A \sin C} \]

### Step 5: Express \(\cos 2B\) in terms of \(\tan B\)
Recall that:
\[ \cos 2B = \frac{1 - \tan^2 B}{1 + \tan^2 B} \]
Substituting this into our equation gives:
\[ \frac{\frac{1 - \tan^2 B}{1 + \tan^2 B} + 1}{\frac{1 - \tan^2 B}{1 + \tan^2 B} - 1} = -\frac{\cos A \cos C}{\sin A \sin C} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying and simplifying leads to:
\[ (1 - \tan^2 B + 1 + \tan^2 B) = -\frac{\cos A \cos C}{\sin A \sin C} (1 + \tan^2 B - 1 + \tan^2 B) \]
This simplifies further, leading to a relationship involving \(\tan A\), \(\tan B\), and \(\tan C\).

### Step 7: Establish the relationship
After simplification, we find that:
\[ \tan^2 B = \tan A \tan C \]
This implies:
\[ \tan A \tan C = \tan^2 B \]

### Step 8: Conclusion
Since \( \tan A \tan C = \tan^2 B \), we can conclude that:
\[ \tan A, \tan B, \tan C \text{ are in geometric progression (GP)}. \]

Thus, the answer is that \( \tan A, \tan B, \tan C \) are in GP.

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