If: `tan^(2) theta + sec 2theta =1`, then the general value of `theta` is

Options

A. `npi`
B. `npi, npi +- pi/3`
C. `npi +- pi/6`
D. none of these

Answer

To solve the equation \( \tan^2 \theta + \sec 2\theta = 1 \), we will follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that:
\[ \sec 2\theta = \frac{1}{\cos 2\theta} \]
Thus, we can rewrite the equation as:
\[ \tan^2 \theta + \frac{1}{\cos 2\theta} = 1 \]

### Step 2: Use the double angle formula for cosine
The double angle formula for cosine states:
\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \]
Substituting this into our equation gives:
\[ \tan^2 \theta + \frac{1}{\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}} = 1 \]

### Step 3: Simplify the equation
This simplifies to:
\[ \tan^2 \theta + \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta} = 1 \]
Now, let's set \( T = \tan^2 \theta \). The equation becomes:
\[ T + \frac{1 + T}{1 - T} = 1 \]

### Step 4: Clear the fraction
Multiply through by \( 1 - T \):
\[ T(1 - T) + (1 + T) = 1 - T \]
Expanding this gives:
\[ T - T^2 + 1 + T = 1 - T \]
Combining like terms results in:
\[ 2T - T^2 + 1 = 1 - T \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ -T^2 + 3T = 0 \]
Factoring out \( T \):
\[ T(-T + 3) = 0 \]

### Step 6: Solve for \( T \)
This gives us two solutions:
1. \( T = 0 \)
2. \( T = 3 \)

### Step 7: Substitute back for \( \tan^2 \theta \)
Since \( T = \tan^2 \theta \):
1. From \( T = 0 \):
\[ \tan^2 \theta = 0 \implies \tan \theta = 0 \implies \theta = n\pi \]
2. From \( T = 3 \):
\[ \tan^2 \theta = 3 \implies \tan \theta = \pm \sqrt{3} \implies \theta = n\pi \pm \frac{\pi}{3} \]

### Step 8: General solution
Thus, the general values of \( \theta \) are:
\[ \theta = n\pi \quad \text{and} \quad \theta = n\pi \pm \frac{\pi}{3} \]

### Final Answer
The general values of \( \theta \) are:
\[ \theta = n\pi \quad \text{and} \quad \theta = n\pi \pm \frac{\pi}{3} \]

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