For any real values of `theta,sqrt((sectheta-1)/(Sectheta+1))`=?

Options

A. `cottheta-"cosec"theta`
B. `sectheta-tantheta`
C. `"cosec"theta-cottheta`
D. `tan theta-sectheta`

Answer

To solve the expression \(\sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}}\), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}} \]

### Step 2: Multiply numerator and denominator by \(\sec \theta - 1\)
To simplify the expression, we multiply both the numerator and the denominator by \(\sec \theta - 1\):
\[ \sqrt{\frac{(\sec \theta - 1)(\sec \theta - 1)}{(\sec \theta + 1)(\sec \theta - 1)}} \]

### Step 3: Simplify the numerator
The numerator becomes:
\[ (\sec \theta - 1)^2 \]

### Step 4: Simplify the denominator
The denominator can be simplified using the difference of squares:
\[ \sec^2 \theta - 1 \]
Using the identity \(\sec^2 \theta - 1 = \tan^2 \theta\), we can rewrite the denominator as:
\[ \tan^2 \theta \]

### Step 5: Rewrite the entire expression
Now we can rewrite the expression:
\[ \sqrt{\frac{(\sec \theta - 1)^2}{\tan^2 \theta}} \]

### Step 6: Simplify the square root
Taking the square root gives us:
\[ \frac{\sec \theta - 1}{\tan \theta} \]

### Step 7: Substitute \(\sec \theta\) and \(\tan \theta\)
Recall that:
\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Substituting these into the expression, we have:
\[ \frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}} = \frac{\frac{1 - \cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{1 - \cos \theta}{\sin \theta} \]

### Step 8: Final expression
Thus, we can conclude that:
\[ \sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}} = \frac{1 - \cos \theta}{\sin \theta} \]