Simplify `(sintheta+sectheta)^(2)+(costheta+cosectheta)^(2)`.

Options

A. `(1+sectheta)^(2)`
B. `(1+cosectheta)^(2)`
C. `1+(cosectheta + sectheta)^(2)`
D. `(1-cosectheta.sectheta)^(2)`

Answer

To simplify the expression \((\sin \theta + \sec \theta)^2 + (\cos \theta + \csc \theta)^2\), we will follow these steps:

### Step 1: Apply the square of a binomial formula
We start by applying the identity \((A + B)^2 = A^2 + B^2 + 2AB\) to both terms in the expression.

1. For \((\sin \theta + \sec \theta)^2\):
\[ (\sin \theta + \sec \theta)^2 = \sin^2 \theta + \sec^2 \theta + 2 \sin \theta \sec \theta \]

2. For \((\cos \theta + \csc \theta)^2\):
\[ (\cos \theta + \csc \theta)^2 = \cos^2 \theta + \csc^2 \theta + 2 \cos \theta \csc \theta \]

### Step 2: Combine the results
Now we combine the results from both expansions:
\[ (\sin \theta + \sec \theta)^2 + (\cos \theta + \csc \theta)^2 = (\sin^2 \theta + \sec^2 \theta + 2 \sin \theta \sec \theta) + (\cos^2 \theta + \csc^2 \theta + 2 \cos \theta \csc \theta) \]

### Step 3: Group like terms
Now we group the terms:
\[ = \sin^2 \theta + \cos^2 \theta + \sec^2 \theta + \csc^2 \theta + 2(\sin \theta \sec \theta + \cos \theta \csc \theta) \]

### Step 4: Use trigonometric identities
We know from trigonometric identities that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Also, we can express \(\sec^2 \theta\) and \(\csc^2 \theta\) as:
\[ \sec^2 \theta = 1 + \tan^2 \theta \quad \text{and} \quad \csc^2 \theta = 1 + \cot^2 \theta \]
Thus,
\[ \sec^2 \theta + \csc^2 \theta = (1 + \tan^2 \theta) + (1 + \cot^2 \theta) = 2 + \tan^2 \theta + \cot^2 \theta \]

### Step 5: Substitute back into the expression
Substituting back, we have:
\[ = 1 + (2 + \tan^2 \theta + \cot^2 \theta) + 2(\sin \theta \sec \theta + \cos \theta \csc \theta) \]
This simplifies to:
\[ = 3 + \tan^2 \theta + \cot^2 \theta + 2(\sin \theta \sec \theta + \cos \theta \csc \theta) \]

### Step 6: Final simplification
Now we can simplify the terms \(2(\sin \theta \sec \theta + \cos \theta \csc \theta)\):
\[ \sin \theta \sec \theta = \frac{\sin \theta}{\cos \theta} = \tan \theta \]
\[ \cos \theta \csc \theta = \frac{\cos \theta}{\sin \theta} = \cot \theta \]
Thus,
\[ 2(\sin \theta \sec \theta + \cos \theta \csc \theta) = 2(\tan \theta + \cot \theta) \]

### Final Result
Putting everything together, we have:
\[ = 3 + \tan^2 \theta + \cot^2 \theta + 2(\tan \theta + \cot \theta) \]

This is the simplified form of the original expression.