`(2sinthetatantheta(1-tantheta)+2sinthetasec^(2)theta)/((1+tantheta)^(2))=?`

Options

A. `(sintheta)/(1+tantheta)`
B. `(2sintheta)/(1+tantheta)`
C. `(2sintheta)/((1+tantheta)^(2))`
D. `(2sintheta)/((1-tantheta)^(2))`

Answer

To solve the expression \((2 \sin \theta \tan \theta (1 - \tan \theta) + 2 \sin \theta \sec^2 \theta) / (1 + \tan \theta)^2\), we will follow these steps:

### Step 1: Rewrite the expression
The given expression is:
\[ \frac{2 \sin \theta \tan \theta (1 - \tan \theta) + 2 \sin \theta \sec^2 \theta}{(1 + \tan \theta)^2} \]

### Step 2: Substitute \(\sec^2 \theta\)
We know that \(\sec^2 \theta = 1 + \tan^2 \theta\). Therefore, we can substitute this into the expression:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
So, the expression becomes:
\[ \frac{2 \sin \theta \tan \theta (1 - \tan \theta) + 2 \sin \theta (1 + \tan^2 \theta)}{(1 + \tan \theta)^2} \]

### Step 3: Expand the numerator
Now, let's expand the numerator:
\[ 2 \sin \theta \tan \theta (1 - \tan \theta) = 2 \sin \theta \tan \theta - 2 \sin \theta \tan^2 \theta \]
And for the other part:
\[ 2 \sin \theta (1 + \tan^2 \theta) = 2 \sin \theta + 2 \sin \theta \tan^2 \theta \]
Combining these, we have:
\[ 2 \sin \theta \tan \theta - 2 \sin \theta \tan^2 \theta + 2 \sin \theta + 2 \sin \theta \tan^2 \theta \]
Notice that \(-2 \sin \theta \tan^2 \theta\) and \(2 \sin \theta \tan^2 \theta\) cancel each other out, leaving us with:
\[ 2 \sin \theta \tan \theta + 2 \sin \theta \]

### Step 4: Factor out common terms
Now we can factor out \(2 \sin \theta\) from the numerator:
\[ 2 \sin \theta (\tan \theta + 1) \]

### Step 5: Substitute back into the expression
Now substituting this back into the expression, we have:
\[ \frac{2 \sin \theta (\tan \theta + 1)}{(1 + \tan \theta)^2} \]

### Step 6: Simplify the expression
Notice that \((\tan \theta + 1)\) is the same as \((1 + \tan \theta)\), thus we can simplify:
\[ \frac{2 \sin \theta (1 + \tan \theta)}{(1 + \tan \theta)^2} \]
This simplifies to:
\[ \frac{2 \sin \theta}{1 + \tan \theta} \]

### Final Answer
Thus, the final result is:
\[ \frac{2 \sin \theta}{1 + \tan \theta} \]

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