The value of `sectheta((1+sin theta)/(costheta)+(costheta)/(1+sintheta))-2tan^(2)theta` is

Options

A. 4
B. 1
C. 2
D. 0

Answer

To solve the expression \( \sec \theta \left( \frac{1 + \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 + \sin \theta} \right) - 2 \tan^2 \theta \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \sec \theta \left( \frac{1 + \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 + \sin \theta} \right) - 2 \tan^2 \theta \]
Recall that \( \sec \theta = \frac{1}{\cos \theta} \) and \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \).

### Step 2: Find a common denominator
For the terms inside the parentheses, we need a common denominator:
\[ \frac{1 + \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 + \sin \theta} \]
The common denominator is \( \cos \theta (1 + \sin \theta) \). Thus, we can rewrite:
\[ \frac{(1 + \sin \theta)^2 + \cos^2 \theta}{\cos \theta (1 + \sin \theta)} \]

### Step 3: Expand the numerator
Now we expand \( (1 + \sin \theta)^2 \):
\[ (1 + \sin \theta)^2 = 1 + 2\sin \theta + \sin^2 \theta \]
So the numerator becomes:
\[ 1 + 2\sin \theta + \sin^2 \theta + \cos^2 \theta \]
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can simplify:
\[ 1 + 2\sin \theta + 1 = 2 + 2\sin \theta \]

### Step 4: Substitute back into the expression
Now we substitute back into our expression:
\[ \sec \theta \cdot \frac{2 + 2\sin \theta}{\cos \theta (1 + \sin \theta)} - 2 \tan^2 \theta \]
This simplifies to:
\[ \frac{2 + 2\sin \theta}{\cos^2 \theta (1 + \sin \theta)} - 2 \frac{\sin^2 \theta}{\cos^2 \theta} \]

### Step 5: Combine the fractions
Now we combine the fractions:
\[ \frac{2 + 2\sin \theta - 2\sin^2 \theta (1 + \sin \theta)}{\cos^2 \theta (1 + \sin \theta)} \]

### Step 6: Simplify the numerator
The numerator simplifies as follows:
\[ 2 + 2\sin \theta - 2\sin^2 \theta - 2\sin^3 \theta \]
Rearranging gives:
\[ 2(1 + \sin \theta - \sin^2 \theta - \sin^3 \theta) \]

### Step 7: Factor the numerator
Notice that \( 1 + \sin \theta - \sin^2 \theta - \sin^3 \theta \) can be factored:
\[ = 2(1 - \sin^2 \theta) = 2\cos^2 \theta \]
Thus, we have:
\[ \frac{2 \cos^2 \theta}{\cos^2 \theta (1 + \sin \theta)} \]

### Step 8: Cancel out terms
The \( \cos^2 \theta \) cancels out:
\[ \frac{2}{1 + \sin \theta} \]

### Step 9: Final expression
Now we compute the value:
\[ \frac{2}{1 + \sin \theta} - 2 \tan^2 \theta \]
Substituting \( \tan^2 \theta \) gives us:
\[ \frac{2}{1 + \sin \theta} - 2 \frac{\sin^2 \theta}{\cos^2 \theta} \]
This leads us to the final answer, which simplifies to \( 2 \).

### Conclusion
Thus, the value of the expression is:
\[ \boxed{2} \]