| Date | May 2012 | Marks available | 6 | Reference code | 12M.1.hl.TZ2.9 |
| Level | HL only | Paper | 1 | Time zone | TZ2 |
| Command term | Show that | Question number | 9 | Adapted from | N/A |
Question
Show that \(\frac{{\cos A + \sin A}}{{\cos A - \sin A}} = \sec 2A + \tan 2A\) .
Markscheme
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]
Examiners report
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.
