MHB 8.aux.27 Simplify the trig expression

[karush]

$\tiny{8.aux.27}$
Simplify the expression
$\dfrac{{\cos 2x\ }}{{\cos x-{\sin x\ }\ }}
=\dfrac{{{\cos}^2 x-{{\sin}^2 x\ }\ }}{{\cos x\ }-{\sin x\ }}
=\dfrac{({\cos x}-{\sin x})({\cos x}+{\sin x\ })}{{\cos x}-{\sin x}}
=\cos x +\sin x$

ok spent an hour just to get this and still not sure
suggestions?

 

[DaalChawal]

it's correct

 

[topsquark]

With the proviso that we only use x s.t. [math]cos(x) \neq sin(x)[/math]. Since the reason for this has left the expression we need to state that.

-Dan

 

[karush]

good point otherwise you get 0/0