If `theta in (pi//4, pi//2)` and `sum_(n=1)^(oo)(1)/(tan^(n)theta)=sin theta + cos theta`, then the value of tan `theta` is

Options

A. `sqrt(3)`
B. `sqrt(2)+1`
C. `2+sqrt(3)`
D. `sqrt(2)`

Explanation

To solve the problem, we need to find the value of \(\tan \theta\) given that

\[ \sum_{n=1}^{\infty} \frac{1}{\tan^n \theta} = \sin \theta + \cos \theta \]

where \(\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)\).

### Step 1: Recognize the series as a geometric series

The series \(\sum_{n=1}^{\infty} \frac{1}{\tan^n \theta}\) can be recognized as a geometric series with the first term \(a = \frac{1}{\tan \theta}\) and the common ratio \(r = \frac{1}{\tan \theta}\). The sum of an infinite geometric series is given by:

\[ S = \frac{a}{1 - r} = \frac{\frac{1}{\tan \theta}}{1 - \frac{1}{\tan \theta}} = \frac{\frac{1}{\tan \theta}}{\frac{\tan \theta - 1}{\tan \theta}} = \frac{1}{\tan \theta - 1} \]

### Step 2: Set the equation

Now we can set this equal to \(\sin \theta + \cos \theta\):

\[ \frac{1}{\tan \theta - 1} = \sin \theta + \cos \theta \]

### Step 3: Cross-multiply

Cross-multiplying gives:

\[ 1 = (\tan \theta - 1)(\sin \theta + \cos \theta) \]

Expanding this yields:

\[ 1 = \tan \theta \sin \theta + \tan \theta \cos \theta - \sin \theta - \cos \theta \]

### Step 4: Substitute \(\tan \theta\)

Recall that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Substituting this into the equation gives:

\[ 1 = \left(\frac{\sin \theta}{\cos \theta}\right) \sin \theta + \left(\frac{\sin \theta}{\cos \theta}\right) \cos \theta - \sin \theta - \cos \theta \]

This simplifies to:

\[ 1 = \frac{\sin^2 \theta}{\cos \theta} + \sin \theta - \sin \theta - \cos \theta \]

Thus, we have:

\[ 1 = \frac{\sin^2 \theta}{\cos \theta} - \cos \theta \]

### Step 5: Multiply through by \(\cos \theta\)

Multiplying through by \(\cos \theta\) gives:

\[ \cos \theta = \sin^2 \theta - \cos^2 \theta \]

### Step 6: Use the identity \(\sin^2 \theta + \cos^2 \theta = 1\)

We can rewrite \(\sin^2 \theta\) as \(1 - \cos^2 \theta\):

\[ \cos \theta = (1 - \cos^2 \theta) - \cos^2 \theta \]

This simplifies to:

\[ \cos \theta = 1 - 2\cos^2 \theta \]

### Step 7: Rearranging the equation

Rearranging gives us:

\[ 2\cos^2 \theta + \cos \theta - 1 = 0 \]

### Step 8: Solve the quadratic equation

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 2\), \(b = 1\), and \(c = -1\):

\[ \cos \theta = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]

This gives:

\[ \cos \theta = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad \cos \theta = \frac{-4}{4} = -1 \]

### Step 9: Determine valid solutions

Since \(\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)\), \(\cos \theta\) must be positive, thus:

\[ \cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3} \]

### Step 10: Find \(\tan \theta\)

Finally, we find \(\tan \theta\):

\[ \tan \theta = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]

Thus, the value of \(\tan \theta\) is:

\[ \boxed{\sqrt{3}} \]