Let t in (0, 1) and alpha in (0, frac{pi}{4}) | vedclass

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(A) Step $1$: Simplify $x$. Let $t = \sin 	heta$. Then $x = 	ext{cosec}^{-1} \left( \frac{1 + \sin^2 	heta}{2 \sin 	heta} ight)$. This does not

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$\sec \alpha$

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(A) Step $1$: Simplify $x$. Let $t = \sin 	heta$. Then $x = 	ext{cosec}^{-1} \left( \frac{1 + \sin^2 	heta}{2 \sin 	heta} ight)$. This does not simplify easily. Let $t = 	an 	heta$. Then $x = 	ext{cosec}^{-1} \left( \frac{1 + 	an^2 	heta}{2 	an 	heta} ight) = 	ext{cosec}^{-1} \left( \frac{\sec^2 	heta}{2 	an 	heta} ight) = 	ext{cosec}^{-1} \left( \frac{1}{2 \sin 	heta \cos 	heta} ight) = 	ext{cosec}^{-1} (	ext{cosec} 2	heta) = 2	heta = 2 	an^{-1} t$.Step $2$: Simplify $y$. Let $t = \sin 	heta$. Then $y = \cot^{-1} \left( \frac{\sqrt{1 - \sin^2 	heta}}{\sin 	heta} ight) = \cot^{-1} \left( \frac{\cos 	heta}{\sin 	heta} ight) = \cot^{-1} (\cot 	heta) = 	heta = \sin^{-1} t$.Step $3$: Find $\frac{dy}{dx}$. $\frac{dx}{dt} = \frac{2}{1 + t^2}$ and $\frac{dy}{dt} = \frac{1}{\sqrt{1 - t^2}}$. Thus, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1}{\sqrt{1 - t^2}} \cdot \frac{1 + t^2}{2} = \frac{1 + t^2}{2 \sqrt{1 - t^2}}$.Step $4$: Calculate $f(	an \alpha)$. $f(	an \alpha) = \frac{1 + 	an^2 \alpha}{2 \sqrt{1 - 	an^2 \alpha}} = \frac{\sec^2 \alpha}{2 \sqrt{1 - 	an^2 \alpha}} = \frac{1}{2 \cos^2 \alpha \sqrt{\frac{\cos^2 \alpha - \sin^2 \alpha}{\cos^2 \alpha}}} = \frac{1}{2 \cos \alpha \sqrt{\cos 2\alpha}}$. Note: Re-evaluating $y = \sin^{-1} t$, $\frac{dy}{dt} = \frac{1}{\sqrt{1-t^2}}$. The derivative is $\frac{1+t^2}{2\sqrt{1-t^2}}$. Given the options, the intended answer is $\frac{1}{2\cos\alpha\sqrt{\cos 2\alpha}}$. Since this is not listed, we re-check $y$. If $y = \sec^{-1}(1/t) = \cos^{-1} t$, then $\frac{dy}{dt} = -\frac{1}{\sqrt{1-t^2}}$. The result depends on the definition of $y$.

Let $t \in (0, 1)$ and $\alpha \in (0, \frac{\pi}{4})$. If $x = \text{cosec}^{-1} \left( \frac{1 + t^2}{2t} \right)$, $y = \cot^{-1} \left( \frac{\sqrt{1 - t^2}}{t} \right)$ and $\frac{dy}{dx} = f(t)$, then the value of $f(\tan \alpha)$ is

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