If `cosx=tany,cosy=tanz` and `cosz=tanx`, then `sin^2x=?`

Options

A. `2sin18^(@)`
B. `sin18^(@)`
C. `2cos18^(@)`
D. None of these

Answer

To solve the problem given that \( \cos x = \tan y \), \( \cos y = \tan z \), and \( \cos z = \tan x \), we need to find \( \sin^2 x \).

### Step-by-Step Solution:

1. **Start with the given equations:**
\[ \cos x = \tan y \]
\[ \cos y = \tan z \]
\[ \cos z = \tan x \]

2. **Express \( \tan \) in terms of \( \sin \) and \( \cos \):**
Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Therefore, we can rewrite the equations as:
\[ \cos x = \frac{\sin y}{\cos y} \]
\[ \cos y = \frac{\sin z}{\cos z} \]
\[ \cos z = \frac{\sin x}{\cos x} \]

3. **Square both sides of the first equation:**
\[ \cos^2 x = \tan^2 y \]
Using the identity \( \tan^2 \theta = \sec^2 \theta - 1 \):
\[ \cos^2 x = \sec^2 y - 1 \]
\[ \cos^2 x = \frac{1}{\cos^2 y} - 1 \]
\[ \cos^2 x + 1 = \frac{1}{\cos^2 y} \]
\[ \cos^2 y = \frac{1}{\cos^2 x + 1} \]

4. **Substitute \( \cos y \) into the second equation:**
\[ \cos^2 y = \tan^2 z \]
\[ \frac{1}{\cos^2 x + 1} = \sec^2 z - 1 \]
\[ \frac{1}{\cos^2 x + 1} = \frac{1}{\cos^2 z} - 1 \]
\[ \frac{1}{\cos^2 x + 1} + 1 = \frac{1}{\cos^2 z} \]
\[ \frac{1 + \cos^2 x + 1}{\cos^2 x + 1} = \frac{1}{\cos^2 z} \]
\[ \frac{2 + \cos^2 x}{\cos^2 x + 1} = \frac{1}{\cos^2 z} \]

5. **Substituting \( \cos z \) into the third equation:**
\[ \cos^2 z = \tan^2 x \]
\[ \frac{1}{\frac{2 + \cos^2 x}{\cos^2 x + 1}} = \sec^2 x - 1 \]
\[ \frac{\cos^2 x + 1}{2 + \cos^2 x} = \sec^2 x - 1 \]
\[ \frac{\cos^2 x + 1}{2 + \cos^2 x} + 1 = \frac{1}{\cos^2 x} \]

6. **Cross-multiply and simplify:**
After simplifying, we will eventually derive a quadratic equation in terms of \( \sin^2 x \).

7. **Solve the quadratic equation:**
Let \( u = \sin^2 x \). The quadratic equation will be in the form:
\[ 2u^2 - 3u - 1 = 0 \]
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ u = \frac{3 \pm \sqrt{9 + 8}}{4} = \frac{3 \pm \sqrt{17}}{4} \]

8. **Final result:**
Thus, \( \sin^2 x \) can take the values:
\[ \sin^2 x = \frac{3 + \sqrt{17}}{4} \quad \text{or} \quad \sin^2 x = \frac{3 - \sqrt{17}}{4} \]