What type of discontinuity does `f(x)=cos^(-1)((1-tan^(2) ""(x)/(2))/(1+tan^(2)""(x)/(2)))` have ?

Answer

To determine the type of discontinuity for the function \( f(x) = \cos^{-1} \left( \frac{1 - \tan^2 \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)} \right) \), we can follow these steps:

### Step 1: Rewrite the function using trigonometric identities
We know that:
\[ \cos(2x) = \frac{1 - \tan^2(x)}{1 + \tan^2(x)} \]
Thus, we can express \( f(x) \) as:
\[ f(x) = \cos^{-1}(\cos(x)) \]

### Step 2: Identify the range of the inverse cosine function
The function \( \cos^{-1}(x) \) has a range of \( [0, \pi] \). However, since \( x \) can take any real value, we need to consider the periodic nature of the cosine function.

### Step 3: Determine points of discontinuity
The function \( \cos^{-1}(\cos(x)) \) is defined for all \( x \) but will have discontinuities at points where \( x \) is an odd multiple of \( \pi \):
\[ x = (2n + 1)\pi, \quad n \in \mathbb{Z} \]
At these points, the function \( \tan\left(\frac{x}{2}\right) \) becomes undefined because:
\[ \tan\left(\frac{(2n + 1)\pi}{2}\right) = \tan\left((n + \frac{1}{2})\pi\right) = \text{undefined} \]

### Step 4: Classify the type of discontinuity
Since the function is defined and continuous everywhere else except at these specific points, where it is undefined, we can classify this as a removable discontinuity. This means that we can redefine the function at these points to make it continuous.

### Final Conclusion
Thus, the function \( f(x) \) has a removable discontinuity at \( x = (2n + 1)\pi \) for \( n \in \mathbb{Z} \).

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