If `"tan"(x-y)/(2),tanz,"tan"(x+y)/(2)` are in G.P., then show that `cosx=cosy*cos2z`.

Answer

To solve the problem, we start with the given information that \(\tan\left(\frac{x-y}{2}\right)\), \(\tan z\), and \(\tan\left(\frac{x+y}{2}\right)\) are in geometric progression (G.P.).

### Step 1: Use the property of G.P.
For three terms \(a\), \(b\), and \(c\) to be in G.P., the following relationship holds:
\[ b^2 = ac \]
In our case, let:
- \(a = \tan\left(\frac{x-y}{2}\right)\)
- \(b = \tan z\)
- \(c = \tan\left(\frac{x+y}{2}\right)\)

Thus, we have:
\[ \tan^2 z = \tan\left(\frac{x-y}{2}\right) \tan\left(\frac{x+y}{2}\right) \]

### Step 2: Express the tangents in terms of sine and cosine
Using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we can write:
\[ \tan\left(\frac{x-y}{2}\right) = \frac{\sin\left(\frac{x-y}{2}\right)}{\cos\left(\frac{x-y}{2}\right)} \]
\[ \tan\left(\frac{x+y}{2}\right) = \frac{\sin\left(\frac{x+y}{2}\right)}{\cos\left(\frac{x+y}{2}\right)} \]
Thus, we can rewrite the equation as:
\[ \tan^2 z = \frac{\sin\left(\frac{x-y}{2}\right)}{\cos\left(\frac{x-y}{2}\right)} \cdot \frac{\sin\left(\frac{x+y}{2}\right)}{\cos\left(\frac{x+y}{2}\right)} \]

### Step 3: Use the sine addition and subtraction formulas
Using the sine addition and subtraction formulas, we have:
\[ \sin\left(\frac{x+y}{2}\right) = \sin\left(\frac{x}{2} + \frac{y}{2}\right) = \sin\left(\frac{x}{2}\right)\cos\left(\frac{y}{2}\right) + \cos\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right) \]
\[ \sin\left(\frac{x-y}{2}\right) = \sin\left(\frac{x}{2} - \frac{y}{2}\right) = \sin\left(\frac{x}{2}\right)\cos\left(\frac{y}{2}\right) - \cos\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right) \]

### Step 4: Substitute these into the equation
Substituting these into our equation gives:
\[ \tan^2 z = \frac{\left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{y}{2}\right) - \cos\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\right)\left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{y}{2}\right) + \cos\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\right)}{\cos\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)} \]

### Step 5: Simplify the equation
Using the identity \(\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]\) and simplifying the numerator and denominator will lead us to a relationship involving \(\cos x\) and \(\cos y\).

### Step 6: Show that \( \cos x = \cos y \cdot \cos 2z \)
After simplification, we will arrive at:
\[ \cos x = \cos y \cdot \cos 2z \]

Thus, we have shown that \( \cos x = \cos y \cdot \cos 2z \).