Show that $\frac{{\cos A + \sin A}}{{\cos A - \sin A}} = \sec 2A + \tan 2A$ .

METHOD 1

$\frac{{\cos A + \sin A}}{{\cos A - \sin A}} = \sec 2A + \tan 2A$

consider right hand side

$\sec 2A + \tan 2A = \frac{1}{{\cos 2A}} + \frac{{\sin 2A}}{{\cos 2A}}$     M1A1

$= \frac{{{{\cos }^2}A + 2\sin A\cos A + {{\sin }^2}A}}{{{{\cos }^2}A - {{\sin }^2}A}}$     A1A1 

Note: Award A1 for recognizing the need for single angles and A1 for recognizing ${\cos ^2}A + {\sin ^2}A = 1$ .

$= \frac{{{{(\cos A + \sin A)}^2}}}{{(\cos A + \sin A)(\cos A - \sin A)}}$     M1A1

$= \frac{{\cos A + \sin A}}{{\cos A - \sin A}}$     AG

METHOD 2

$\frac{{\cos A + \sin A}}{{\cos A - \sin A}} = \frac{{{{(\cos A + \sin A)}^2}}}{{(\cos A + \sin A)(\cos A - \sin A)}}$     M1A1

$= \frac{{{{\cos }^2}A + 2\sin A\cos A + {{\sin }^2}A}}{{{{\cos }^2}A - {{\sin }^2}A}}$     A1A1 

Note: Award A1 for correct numerator and A1 for correct denominator.

$= \frac{{1 + \sin 2A}}{{\cos 2A}}$     M1A1

$= \sec 2A + \tan 2A$     AG

[6 marks]

Solutions to this question were good in general with many candidates realising that multiplying the numerator and denominator by $(\cos A + \sin A)$ might be helpful.