Let $f(\theta) = \sin\left(\tan^{-1}\left(\frac{\sin\theta}{\sqrt{\cos 2\theta}}\right)\right)$, where $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$. Then the value of $\frac{d}{d(\tan\theta)}(f(\theta))$ is ______.

Let f(θ) = sin(tan⁻¹(sinθ/(√cos 2θ))), where -π/4 < θ < π/4. Then the value of d/d(tanθ)(f(θ)) is ______.

$f(\theta) = \sin\left(\tan^{-1}\left(\dfrac{\sin\theta}{\sqrt{\cos 2\theta}}\right)\right)$

$-\dfrac{\pi}{4} < \theta < \dfrac{\pi}{4}$

$\text{To find: } \dfrac{d\,f(\theta)}{d(\tan\theta)}$

$\text{Let } \phi = \tan^{-1}\!\left(\dfrac{\sin\theta}{\sqrt{\cos 2\theta}}\right)$

$\Rightarrow \tan\phi = \dfrac{\sin\theta}{\sqrt{\cos 2\theta}}, \quad f = \sin\phi$

$\sin\phi = \dfrac{\tan\phi}{\sqrt{1+\tan^2\phi}}$

$\sin^2\phi = \dfrac{\sin^2\theta/\cos 2\theta}{1+\sin^2\theta/\cos 2\theta} = \dfrac{\sin^2\theta}{\cos 2\theta + \sin^2\theta}$

$\cos 2\theta + \sin^2\theta = \cos^2\theta$

$\sin^2\phi = \tan^2\theta \Rightarrow \sin\phi = \tan\theta$

$f(\theta) = \sin\phi = \tan\theta$

$\dfrac{d\,f(\theta)}{d(\tan\theta)} = 1$

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