`tan^(-1)(sinx)=sin^(-1)(tanx)` holds true for (A) `"x" epsilonR` (B) `2npi-(pi)/2lexle2npi+(pi)/2("n" epsilonz)` (C) `"x" epsilon{0,z^(+)}` (D) `"x" epsilonnpi("n" epsilonz)`

Options

A. `"x" epsilonR`
B. `2npi-(pi)/2lexle2npi+(pi)/2("n" epsilonz)`
C. `"x" epsilon{0,z^(+)}`
D. `"x" epsilonnpi("n" epsilonz)`

Explanation

To solve the equation \( \tan^{-1}(\sin x) = \sin^{-1}(\tan x) \), we will follow a step-by-step approach.

### Step 1: Rewrite the equation
We start with the given equation:
\[ \tan^{-1}(\sin x) = \sin^{-1}(\tan x) \]

### Step 2: Use the identity for \(\sin^{-1}\)
We can use the identity for \(\sin^{-1}(\theta)\):
\[ \sin^{-1}(\theta) = \tan^{-1}\left(\frac{\theta}{\sqrt{1 - \theta^2}}\right) \]
Applying this to the right-hand side:
\[ \sin^{-1}(\tan x) = \tan^{-1}\left(\frac{\tan x}{\sqrt{1 - \tan^2 x}}\right) \]

### Step 3: Substitute into the equation
Now we can rewrite the equation as:
\[ \tan^{-1}(\sin x) = \tan^{-1}\left(\frac{\tan x}{\sqrt{1 - \tan^2 x}}\right) \]

### Step 4: Set the arguments equal
Since both sides are \(\tan^{-1}\), we can set the arguments equal to each other:
\[ \sin x = \frac{\tan x}{\sqrt{1 - \tan^2 x}} \]

### Step 5: Rewrite \(\tan x\)
Recall that \(\tan x = \frac{\sin x}{\cos x}\). Substituting this into the equation gives:
\[ \sin x = \frac{\frac{\sin x}{\cos x}}{\sqrt{1 - \left(\frac{\sin x}{\cos x}\right)^2}} \]

### Step 6: Simplify the right-hand side
The denominator simplifies as follows:
\[ \sqrt{1 - \tan^2 x} = \sqrt{1 - \frac{\sin^2 x}{\cos^2 x}} = \sqrt{\frac{\cos^2 x - \sin^2 x}{\cos^2 x}} = \frac{\sqrt{\cos^2 x - \sin^2 x}}{\cos x} \]
Thus, we have:
\[ \sin x = \frac{\frac{\sin x}{\cos x}}{\frac{\sqrt{\cos^2 x - \sin^2 x}}{\cos x}} = \frac{\sin x}{\sqrt{\cos^2 x - \sin^2 x}} \]

### Step 7: Cross-multiply
Cross-multiplying gives:
\[ \sin x \sqrt{\cos^2 x - \sin^2 x} = \sin x \]

### Step 8: Factor out \(\sin x\)
This leads to two cases:
1. \(\sin x = 0\)
2. \(\sqrt{\cos^2 x - \sin^2 x} = 1\)

### Step 9: Solve the first case
From \(\sin x = 0\), we have:
\[ x = n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 10: Solve the second case
From \(\sqrt{\cos^2 x - \sin^2 x} = 1\), squaring both sides gives:
\[ \cos^2 x - \sin^2 x = 1 \]
This simplifies to:
\[ \cos^2 x = 1 \quad \Rightarrow \quad x = n\pi \quad \text{for } n \in \mathbb{Z} \]

### Conclusion
Both cases lead to the same solution:
\[ x = n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus, the correct option is (D) \( x \in n\pi \) (where \( n \in \mathbb{Z} \)).