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Inverse Sine of a Tangent Expression
## Formula
If:
$$\theta=\tan^{-1}x$$
then:
$$\sin\theta=\frac{x}{\sqrt{1+x^2}}$$
Therefore:
$$\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)=\tan^{-1}x$$
## Usage
- Used to convert an inverse sine expression into inverse tangent form.
