User interface language: English | Español

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.

 

Syllabus sections

Topic 3 - Core: Circular functions and trigonometry » 3.3 » Compound angle identities.

View options