Question
Let $$0 < x < \frac{\pi }{4}\,{\text{then}}\left( {\sec 2x - \tan 2x} \right)$$ equals
A.
$$\tan \left( {x - \frac{\pi }{4}} \right)$$
B.
$$\tan \left( {\frac{\pi }{4} - x} \right)$$
C.
$$\tan \left( {x + \frac{\pi }{4}} \right)$$
D.
$${\tan ^2}\left( {x + \frac{\pi }{4}} \right)$$
Answer :
$$\tan \left( {\frac{\pi }{4} - x} \right)$$
Solution :
$$\eqalign{
& \sec 2x - \tan 2x \cr
& = \,\frac{{1 - \sin 2x}}{{\cos 2x}} \cr
& = \frac{{1 - \cos 2\left( {\frac{\pi }{4} - x} \right)}}{{\sin 2\left( {\frac{\pi }{4} - x} \right)}} \cr
& = \frac{{2{{\sin }^2}\left( {\frac{\pi }{4} - x} \right)}}{{2\sin \left( {\frac{\pi }{4} - x} \right)\cos \left( {\frac{\pi }{4} - x} \right)}} \cr
& = \tan \left( {\frac{\pi }{4} - x} \right) \cr} $$