What is `((sin theta +cos theta)(tan theta +cot theta))/(sec theta + cosec theta)` equal to ?

Options

A. 1
B. 2
C. `sin theta`
D. `cos theta`

Answer

To solve the expression \(\frac{(\sin \theta + \cos \theta)(\tan \theta + \cot \theta)}{\sec \theta + \csc \theta}\), we will break it down step by step.

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{(\sin \theta + \cos \theta)(\tan \theta + \cot \theta)}{\sec \theta + \csc \theta} \]

### Step 2: Substitute \(\tan \theta\) and \(\cot \theta\)
Recall that:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta} \]
Thus, we can rewrite \(\tan \theta + \cot \theta\) as:
\[ \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \]
Finding a common denominator:
\[ \tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \]
Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ \tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta} \]

### Step 3: Substitute \(\sec \theta\) and \(\csc \theta\)
Recall that:
\[ \sec \theta = \frac{1}{\cos \theta}, \quad \csc \theta = \frac{1}{\sin \theta} \]
Thus, we can rewrite \(\sec \theta + \csc \theta\) as:
\[ \sec \theta + \csc \theta = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \]
Finding a common denominator:
\[ \sec \theta + \csc \theta = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} \]

### Step 4: Substitute back into the expression
Now substituting back into our original expression:
\[ \frac{(\sin \theta + \cos \theta) \cdot \frac{1}{\sin \theta \cos \theta}}{\frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta}} \]

### Step 5: Simplify the expression
The \(\sin \theta + \cos \theta\) terms in the numerator and denominator cancel out:
\[ = \frac{1}{1} = 1 \]

### Final Result
Thus, the value of the expression is:
\[ \boxed{1} \]