DIfferential coefficient of `tan^-1(x/(1+sqrt(1-x^2)))w.r.t sin^-1x` is

Options

A. `1/2`
B. 1
C. 2
D. `3/2`

Answer

To find the differential coefficient of \( \tan^{-1} \left( \frac{x}{1 + \sqrt{1 - x^2}} \right) \) with respect to \( \sin^{-1}(x) \), we will follow these steps:

### Step 1: Define the variables
Let:
\[ y = \tan^{-1} \left( \frac{x}{1 + \sqrt{1 - x^2}} \right) \]
\[ z = \sin^{-1}(x) \]

### Step 2: Express \( x \) in terms of \( z \)
Since \( z = \sin^{-1}(x) \), we can express \( x \) as:
\[ x = \sin(z) \]

### Step 3: Substitute \( x \) into \( y \)
Now substitute \( x = \sin(z) \) into \( y \):
\[ y = \tan^{-1} \left( \frac{\sin(z)}{1 + \sqrt{1 - \sin^2(z)}} \right) \]
Using the identity \( \sqrt{1 - \sin^2(z)} = \cos(z) \), we have:
\[ y = \tan^{-1} \left( \frac{\sin(z)}{1 + \cos(z)} \right) \]

### Step 4: Simplify the expression
Using the half-angle identity, we know that:
\[ 1 + \cos(z) = 2 \cos^2\left(\frac{z}{2}\right) \]
Thus, we can rewrite \( y \):
\[ y = \tan^{-1} \left( \frac{\sin(z)}{2 \cos^2\left(\frac{z}{2}\right)} \right) \]

### Step 5: Use the tangent half-angle formula
Using the tangent half-angle formula:
\[ \tan\left(\frac{z}{2}\right) = \frac{\sin(z)}{1 + \cos(z)} \]
we can simplify \( y \):
\[ y = \tan^{-1} \left( \tan\left(\frac{z}{2}\right) \right) = \frac{z}{2} \]

### Step 6: Differentiate \( y \) with respect to \( z \)
Now we differentiate \( y \) with respect to \( z \):
\[ \frac{dy}{dz} = \frac{1}{2} \]

### Step 7: Find the differential coefficient
The differential coefficient of \( y \) with respect to \( z \) is:
\[ \frac{dy}{dz} = \frac{1}{2} \]

Thus, the final answer is:
\[ \frac{dy}{dz} = \frac{1}{2} \]