Express `sin^(-1).(sqrtx)/(sqrt(x + a))` as a function of `tan^(-1)`

Explanation

To express \( \sin^{-1}\left(\frac{\sqrt{x}}{\sqrt{x + a}}\right) \) as a function of \( \tan^{-1} \), we can follow these steps:

### Step 1: Substitute \( x \)
Let \( x = a \tan^2 \theta \). This substitution will help us simplify the expression.

### Step 2: Rewrite the expression
Now, substitute \( x \) in the original expression:
\[ \sin^{-1}\left(\frac{\sqrt{x}}{\sqrt{x + a}}\right) = \sin^{-1}\left(\frac{\sqrt{a \tan^2 \theta}}{\sqrt{a \tan^2 \theta + a}}\right) \]

### Step 3: Simplify the numerator and denominator
The expression simplifies to:
\[ \sin^{-1}\left(\frac{\sqrt{a} \tan \theta}{\sqrt{a(\tan^2 \theta + 1)}}\right) \]
Since \( \tan^2 \theta + 1 = \sec^2 \theta \), we can rewrite the denominator:
\[ \sin^{-1}\left(\frac{\sqrt{a} \tan \theta}{\sqrt{a} \sec \theta}\right) \]

### Step 4: Cancel out \( \sqrt{a} \)
This simplifies to:
\[ \sin^{-1}\left(\frac{\tan \theta}{\sec \theta}\right) = \sin^{-1}(\sin \theta) \]

### Step 5: Simplify further
Since \( \sin^{-1}(\sin \theta) = \theta \), we have:
\[ \theta = \tan^{-1}(\tan \theta) \]

### Step 6: Relate \( \theta \) back to \( x \)
From our substitution \( x = a \tan^2 \theta \), we can express \( \tan \theta \) as:
\[ \tan \theta = \sqrt{\frac{x}{a}} \]
Thus,
\[ \theta = \tan^{-1}\left(\frac{\sqrt{x}}{\sqrt{a}}\right) \]

### Final Result
So, we can express \( \sin^{-1}\left(\frac{\sqrt{x}}{\sqrt{x + a}}\right) \) as:
\[ \sin^{-1}\left(\frac{\sqrt{x}}{\sqrt{x + a}}\right) = \tan^{-1}\left(\frac{\sqrt{x}}{\sqrt{a}}\right) \]

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