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**Tweety** $\displaystyle \frac{dx}{dy} = \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2} $

$\displaystyle \frac{dy}{dx} = \frac{1}{ \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}}

$

$\displaystyle sec y = 2x $ so replace $\displaystyle \frac{secy}{2} $ with 2x

and using the indentity $\displaystyle 1 + tan^{2}y = sec^{2}y $

$\displaystyle sec^{2}y - 1 = tan^{2}y $

$\displaystyle (\frac{tany}{2})^{2} = \frac{sec^{2}y -1}{4} $

$\displaystyle 2x = secy $

$\displaystyle (2x)^{2} = sec^{2}y $

$\displaystyle 4x^{2} = sec^{2}y $

Can I rearrange and replace it like this?

So I'll get, ;

$\displaystyle \frac{1}{ \frac{1}{2}\times \frac{2x}{2} \times \frac{sec^{2}y-1}{4}} $

cancel the two's and replace sec^2y

$\displaystyle \frac{1}{\frac{1}{2} \times x \times \frac{4x^{2}-1}{4}} $

so after simplifying I get, $\displaystyle \frac{1}{\frac{x}{2} \times \frac{4x^{2}-1}{4}} $

Can someone please help me from here, as I have tired to rearrange it but don't get the desired expression.

And I am not even sure what I have done so far is correct, which is probably why I am not getting the right expression?

Please check my working,

Thank you!