If `cos x=tany, cos y=tanz and cos z = tanx`, then `sinx = 2sin theta` where `theta` is (where, `x,y, z, theta` are acuate angles)

Options

A. `15^(@)`
B. `18^(@)`
C. `22(1^(@))/(2)`
D. `75^(@)`

Answer

To solve the problem, we need to find the value of \( \theta \) such that \( \sin x = 2 \sin \theta \), given the relationships between \( x, y, z \) in terms of trigonometric functions. Let's break this down step by step.

### Step 1: Start with the given equations
We are given:
1. \( \cos x = \tan y \)
2. \( \cos y = \tan z \)
3. \( \cos z = \tan x \)

### Step 2: Square the first equation
From \( \cos x = \tan y \), we square both sides:
\[ \cos^2 x = \tan^2 y \]

### Step 3: Use the identity for tangent
Recall the identity \( \tan^2 y = \sec^2 y - 1 \):
\[ \cos^2 x = \sec^2 y - 1 \]
This implies:
\[ \sec^2 y = \cos^2 x + 1 \]

### Step 4: Rewrite secant in terms of cosine
Since \( \sec y = \frac{1}{\cos y} \), we can write:
\[ \frac{1}{\cos^2 y} = \cos^2 x + 1 \]
Taking the reciprocal gives us:
\[ \cos^2 y = \frac{1}{\cos^2 x + 1} \]

### Step 5: Proceed with the second equation
Now, using \( \cos y = \tan z \) and squaring both sides:
\[ \cos^2 y = \tan^2 z \]
Using \( \tan^2 z = \sec^2 z - 1 \):
\[ \cos^2 y = \sec^2 z - 1 \]
This implies:
\[ \sec^2 z = \cos^2 y + 1 \]

### Step 6: Substitute for \( \cos^2 y \)
Substituting \( \cos^2 y = \frac{1}{\cos^2 x + 1} \):
\[ \sec^2 z = \frac{1}{\cos^2 x + 1} + 1 \]
Taking the reciprocal gives:
\[ \cos^2 z = \frac{1}{\sec^2 z} = \frac{\cos^2 x + 1}{2} \]

### Step 7: Use the third equation
From \( \cos z = \tan x \):
\[ \cos^2 z = \tan^2 x \]
Using \( \tan^2 x = \sec^2 x - 1 \):
\[ \cos^2 z = \sec^2 x - 1 \]

### Step 8: Equate the two expressions for \( \cos^2 z \)
Now we have two expressions for \( \cos^2 z \):
\[ \frac{\cos^2 x + 1}{2} = \sec^2 x - 1 \]

### Step 9: Solve for \( \cos^2 x \)
Multiplying through by 2:
\[ \cos^2 x + 1 = 2\sec^2 x - 2 \]
Substituting \( \sec^2 x = \frac{1}{\cos^2 x} \):
\[ \cos^2 x + 1 = \frac{2}{\cos^2 x} - 2 \]
Multiplying through by \( \cos^2 x \):
\[ \cos^4 x + \cos^2 x = 2 - 2\cos^2 x \]
Rearranging gives:
\[ \cos^4 x + 3\cos^2 x - 2 = 0 \]

### Step 10: Let \( u = \cos^2 x \)
Let \( u = \cos^2 x \):
\[ u^2 + 3u - 2 = 0 \]

### Step 11: Use the quadratic formula
Using the quadratic formula:
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{9 + 8}}{2} = \frac{-3 \pm \sqrt{17}}{2} \]
Since \( u \) must be non-negative, we take:
\[ u = \frac{-3 + \sqrt{17}}{2} \]

### Step 12: Find \( \sin x \)
Since \( \sin^2 x + \cos^2 x = 1 \):
\[ \sin^2 x = 1 - u = 1 - \frac{-3 + \sqrt{17}}{2} = \frac{5 - \sqrt{17}}{2} \]
Thus:
\[ \sin x = \sqrt{\frac{5 - \sqrt{17}}{2}} \]

### Step 13: Relate to \( 2 \sin \theta \)
We need to find \( \theta \) such that:
\[ \sin x = 2 \sin \theta \]
This implies:
\[ \sin \theta = \frac{1}{2} \sin x \]

### Step 14: Determine \( \theta \)
Using the known value \( \sin 18^\circ = \frac{1}{2} \), we conclude:
\[ \theta = 18^\circ \]

### Final Answer
Thus, the value of \( \theta \) is:
\[ \theta = 18^\circ \]