Question

$\frac{\sin x-\cos x+1}{\sin x-\cos x-1}=\sec x-\tan x$. If this is true for all $x$, enter 1, else enter 0.

Answer 1

$L.H.S=\frac{\sin x-\cos x+1}{\sin x-\cos x-1}$
$=\frac{\sin x-\cos x+1}{\sin x-\cos x-1}.\frac{\sin x-\cos x+1}{\sin x-\cos x+1}$
$=\frac{(\sin x-\cos x+1{)}^2}{(\sin x-\cos x{)}^2-1}=\frac{{\sin }^{2}x+{\cos }^{2}x+1-2\sin x\cos x-2\cos x+2\sin x}{1-2\sin x\cos x-1}$
$=\frac{1-\sin x\cos x-\cos x+\sin x}{-\sin x\cos x}=co\sec x-\tan x-co\sec x\sec x+1\ne R.H.S$