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if y =tanx+secx prove that y2=cosx/(1-sinx)²?

$Given: y = \tan x + \sec x \Rightarrow y_{1} = \frac{d y}{d x} = \sec^{2} x + \sec x + \tan x = \sec x (\tan x + \sec x) \Rightarrow y_{2} = \frac{d^{2} y}{d x} = d (\frac{d y}{d x}) = \sec x + \tan x (\tan x + \sec x) + \sec x (\sec x (\tan x + \sec x)) = \sec x {(\tan x + \sec x)}^{2} = \frac{1}{\cos x} (\frac{1 + \sin x}{\cos x})^{2} = \frac{(1 + \sin x)^{2}}{\cos x \cos^{2} x} = \frac{(1 + \sin x)^{2}}{\cos x (1 - \sin^{2} x)} = \frac{{(1 + \sin x)}^{2}}{(\cos x) (1 + \sin x) (1 - \sin x)} = \frac{(1 + \sin x) (1 - \sin x)}{\cos x (1 - \sin x) (1 - \sin x)} = \frac{1 - \sin^{2} x}{\cos x (1 - \sin x)^{2}} = \frac{\cos^{2} x}{\cos x (1 - \sin x)^{2}} = \frac{\cos x}{(1 - \sin x)^{2}}$

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