Question

If $\sec x+\tan x=0$ then prove that
$\frac{d^{2}y}{d{x}^2}=\frac{\cos x}{{(1-\sin x)}^2}$

Answer 1

$y=\sec x+\tan x$

$y=\frac{1+\sin x}{\cos x}$

$\frac{dy}{dx}=\frac{\cos x\frac{d}{dx}(1+\sin x)-(1+\sin x)\frac{d}{dx}\left(\cos x\right)}{{\cos }^{2}x}$

$=\frac{{\cos }^{2}x-(1+\sin x)(-\sin x)}{{\cos }^{2}x}$

$=\frac{{\cos }^{2}x+\sin x+{\sin }^{2}x}{{\cos }^{2}x}$

$=\frac{1+\sin x}{{\cos }^{2}x}$

$=\frac{1+\sin x}{1-{\sin }^{2}x}$

$=\frac{1+\sin x}{(1+\sin x)(1-\sin x)}$

$=\frac{1}{1-\sin x}$

$\frac{d^{2}y}{d{x}^2}=\frac{(1-\sin x)\frac{d}{dx}\left(1\right)-1\frac{d}{dx}(1-\sin x)}{{(1-\sin x)}^2}$

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