Derivative of Composite Inverse Trigonometric Function with respect to Tan Theta

To find the value of $\frac{d}{d(\tan \theta)}(f(\theta))$, we first simplify the given function $f(\theta)$.

Given:
$f(\theta) = \sin \left( \tan^{-1} \left( \frac{\sin \theta}{\sqrt{\cos 2 \theta}} \right) \right)$
where $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$.

Let $x = \frac{\sin \theta}{\sqrt{\cos 2 \theta}}$. Since $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, we have $\cos 2\theta > 0$ and $\cos \theta > 0$.

Using the identity $\sin(\tan^{-1} x) = \frac{x}{\sqrt{1 + x^2}}$, we evaluate $1 + x^2$:
$1 + x^2 = 1 + \left(\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)^2 = 1 + \frac{\sin^2 \theta}{\cos 2 \theta}$

Using the double angle formula $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$:
$1 + x^2 = \frac{\cos 2 \theta + \sin^2 \theta}{\cos 2 \theta} = \frac{(\cos^2 \theta - \sin^2 \theta) + \sin^2 \theta}{\cos 2 \theta} = \frac{\cos^2 \theta}{\cos 2 \theta}$

Taking the square root on both sides:
$\sqrt{1 + x^2} = \sqrt{\frac{\cos^2 \theta}{\cos 2 \theta}} = \frac{|\cos \theta|}{\sqrt{\cos 2 \theta}}$

Since $\cos \theta > 0$ for $\theta \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$, we have $|\cos \theta| = \cos \theta$. Thus:
$\sqrt{1 + x^2} = \frac{\cos \theta}{\sqrt{\cos 2 \theta}}$

Now, substituting $x$ and $\sqrt{1 + x^2}$ back into $f(\theta)$:
$f(\theta) = \frac{x}{\sqrt{1 + x^2}} = \frac{\frac{\sin \theta}{\sqrt{\cos 2 \theta}}}{\frac{\cos \theta}{\sqrt{\cos 2 \theta}}} = \frac{\sin \theta}{\cos \theta} = \tan \theta$

Finally, we find the derivative of $f(\theta)$ with respect to $\tan \theta$:
$\frac{d}{d(\tan \theta)}(f(\theta)) = \frac{d}{d(\tan \theta)}(\tan \theta) = 1$