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If $\mathrm{y}=\sqrt{\tan \mathrm{x}+\sqrt{\tan \mathrm{x}+\sqrt{\tan \mathrm{x}+\ldots \operatorname{tos} \infty}}}$, prove that $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sec ^{2} \mathrm{x}}{2 \mathrm{y}-1}$


$y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\cdots \text { to } \infty}}}$

$y=\sqrt{\tan x+y}$

On squaring both sides,

$y^{2}=\tan x+y$

Differentiating both sides with respect to $x$,

$2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x}$

$\frac{d y}{d x}(2 y-1)=\sec ^{2} x$

$\frac{d y}{d x}=\frac{\sec ^{2} x}{(2 y-1)}$

Hence proved.

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