If `alpha,-(pi)/(2) lt alpha lt (pi)/(2)` is the solution of `4cos theta+5sin theta=1` ,then the value of `tan alpha` is

Options

A. `(sqrt10+10)/12`
B. `(sqrt10-10)/12`
C. `(10-sqrt10)/12`
D. `(sqrt10+10)/6`

Answer

To solve the equation \(4 \cos \theta + 5 \sin \theta = 1\) for \(\tan \alpha\) where \(\alpha\) is a solution in the range \(-\frac{\pi}{2} < \alpha < \frac{\pi}{2}\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 4 \cos \theta + 5 \sin \theta = 1 \]

### Step 2: Express \(\cos \theta\) and \(\sin \theta\) in terms of \(\tan \frac{\theta}{2}\)
Using the half-angle formulas:
- \(\cos \theta = \frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}\)
- \(\sin \theta = \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}\)

Let \(x = \tan \frac{\theta}{2}\). Then we can substitute these into the equation:
\[ 4 \left(\frac{1 - x^2}{1 + x^2}\right) + 5 \left(\frac{2x}{1 + x^2}\right) = 1 \]

### Step 3: Combine the fractions
To combine the fractions, we multiply through by \(1 + x^2\):
\[ 4(1 - x^2) + 10x = 1 + x^2 \]

### Step 4: Rearrange the equation
Expanding and rearranging gives:
\[ 4 - 4x^2 + 10x = 1 + x^2 \]
\[ -5x^2 + 10x + 3 = 0 \]

### Step 5: Multiply by -1
To simplify, we can multiply the entire equation by -1:
\[ 5x^2 - 10x - 3 = 0 \]

### Step 6: Use the quadratic formula
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- Here, \(a = 5\), \(b = -10\), and \(c = -3\).
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 5 \cdot (-3)}}{2 \cdot 5} \]
\[ x = \frac{10 \pm \sqrt{100 + 60}}{10} \]
\[ x = \frac{10 \pm \sqrt{160}}{10} \]
\[ x = \frac{10 \pm 4\sqrt{10}}{10} \]
\[ x = 1 \pm \frac{2\sqrt{10}}{5} \]

### Step 7: Choose the correct root
Since \(\alpha\) is in the range \(-\frac{\pi}{2} < \alpha < \frac{\pi}{2}\), we take the positive root:
\[ \tan \frac{\theta}{2} = 1 - \frac{2\sqrt{10}}{5} \]

### Step 8: Find \(\tan \alpha\)
Using the double angle formula for tangent:
\[ \tan \alpha = \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}} \]
Substituting \(\tan \frac{\theta}{2}\):
\[ \tan \alpha = \frac{2(1 - \frac{2\sqrt{10}}{5})}{1 - (1 - \frac{2\sqrt{10}}{5})^2} \]

### Step 9: Simplify the expression
Calculating the denominator:
\[ (1 - \frac{2\sqrt{10}}{5})^2 = 1 - \frac{4\sqrt{10}}{5} + \frac{40}{25} = 1 - \frac{4\sqrt{10}}{5} + \frac{8}{5} \]
\[ = \frac{5 - 4\sqrt{10} + 8}{5} = \frac{13 - 4\sqrt{10}}{5} \]
Thus:
\[ \tan \alpha = \frac{2(1 - \frac{2\sqrt{10}}{5})}{\frac{13 - 4\sqrt{10}}{5}} = \frac{10(1 - \frac{2\sqrt{10}}{5})}{13 - 4\sqrt{10}} \]

### Final Step: Calculate and simplify
After simplifying, we find the value of \(\tan \alpha\).