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If the transformation $z=\log \tan \frac{x}{2}$ reduces the differential equation $\frac{d^2 y}{d x^2}+\cot x \frac{d y}{d x}+4 y \operatorname{cosec}{ }^2 x=0$ into the form $\frac{d^2 y}{d z^2}+k y=0$ then $k$ is equal to
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4
$\frac{d^2 y}{d z^2}=\frac{d}{d z}\left(\frac{d y}{d z}\right)=\frac{d}{d z}\left(\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\right)$
$=\frac{d}{d z}\left(\frac{\frac{d y}{d x}}{\frac{1}{\sin x}}\right) \quad\left(\because \frac{d z}{d x}=\frac{1}{\sin x}\right)$
$=\frac{d}{d z}\left(\sin x \cdot \frac{d y}{d x}\right)=\frac{\frac{d}{d x}\left(\sin x \cdot \frac{d y}{d x}\right)}{\frac{d z}{d x}}$
$=\sin x\left[\cos x \cdot \frac{d y}{d x}+\sin x \frac{d^2 y}{d x^2}\right]$
$\because \frac{d^2 y}{d z^2}+k y=0$
$\therefore \sin x \cdot \cos x \cdot \frac{d y}{d x}+\sin ^2 x \frac{d^2 y}{d x^2}+k y=0$
Dividing the equation by $\sin ^2 x$, we get $\frac{d^2 y}{d x^2}+\cot x \cdot \frac{d y}{d x}+k \cdot \operatorname{cosec}^2 x \cdot y=0$
Comparing with the equation
$\frac{d^2 y}{d x^2}+\cot x \cdot \frac{d y}{d x}+4 \operatorname{cosec}^2 x \cdot y=0$
We get $k=4$
$=\frac{d}{d z}\left(\frac{\frac{d y}{d x}}{\frac{1}{\sin x}}\right) \quad\left(\because \frac{d z}{d x}=\frac{1}{\sin x}\right)$
$=\frac{d}{d z}\left(\sin x \cdot \frac{d y}{d x}\right)=\frac{\frac{d}{d x}\left(\sin x \cdot \frac{d y}{d x}\right)}{\frac{d z}{d x}}$
$=\sin x\left[\cos x \cdot \frac{d y}{d x}+\sin x \frac{d^2 y}{d x^2}\right]$
$\because \frac{d^2 y}{d z^2}+k y=0$
$\therefore \sin x \cdot \cos x \cdot \frac{d y}{d x}+\sin ^2 x \frac{d^2 y}{d x^2}+k y=0$
Dividing the equation by $\sin ^2 x$, we get $\frac{d^2 y}{d x^2}+\cot x \cdot \frac{d y}{d x}+k \cdot \operatorname{cosec}^2 x \cdot y=0$
Comparing with the equation
$\frac{d^2 y}{d x^2}+\cot x \cdot \frac{d y}{d x}+4 \operatorname{cosec}^2 x \cdot y=0$
We get $k=4$
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