`tan^(-1)[x/sqrt(a^(2)-x^(2))]=`

Options

A. `1/a.sin^(-1)(x/a)`
B. `a.sin^(-1)(x/a)`
C. `sin^(-1)(x/a)`
D. `sin^(-1)(a/x)`

Answer

To solve the equation \( \tan^{-1}\left(\frac{x}{\sqrt{a^2 - x^2}}\right) \), we can follow these steps:

### Step 1: Substitute \( x \) with \( a \sin \theta \)
Let \( x = a \sin \theta \). This substitution will help simplify the expression.

### Step 2: Rewrite the expression
Now, we need to find \( \sqrt{a^2 - x^2} \):
\[ \sqrt{a^2 - x^2} = \sqrt{a^2 - (a \sin \theta)^2} = \sqrt{a^2 - a^2 \sin^2 \theta} = \sqrt{a^2(1 - \sin^2 \theta)} = \sqrt{a^2 \cos^2 \theta} = a \cos \theta \]

### Step 3: Substitute back into the arctangent function
Now substitute \( x \) and \( \sqrt{a^2 - x^2} \) back into the original expression:
\[ \tan^{-1}\left(\frac{x}{\sqrt{a^2 - x^2}}\right) = \tan^{-1}\left(\frac{a \sin \theta}{a \cos \theta}\right) \]

### Step 4: Simplify the expression
The \( a \) in the numerator and denominator cancels out:
\[ \tan^{-1}(\tan \theta) \]

### Step 5: Apply the inverse tangent property
Since \( \tan^{-1}(\tan \theta) = \theta \) (for \( \theta \) in the range of \( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \)):
\[ \tan^{-1}(\tan \theta) = \theta \]

### Step 6: Substitute back for \( \theta \)
Recall that \( \theta = \sin^{-1}\left(\frac{x}{a}\right) \):
\[ \tan^{-1}\left(\frac{x}{\sqrt{a^2 - x^2}}\right) = \sin^{-1}\left(\frac{x}{a}\right) \]

### Final Result
Thus, the final result is:
\[ \tan^{-1}\left(\frac{x}{\sqrt{a^2 - x^2}}\right) = \sin^{-1}\left(\frac{x}{a}\right) \]

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