21M.1.AHL.TZ1.6

METHOD 1

attempt to use a right angled triangle        M1

correct placement of all three values and $\theta$ seen in the triangle        (A1)

$\text{cot} \theta <0$ (since $\text{cosec} \theta >0$ puts $\theta$ in the second quadrant)        R1

$\text{cot} \theta =-\frac{\sqrt{5}}{2}$        A1

Note: Award M1A1R0A0 for $\text{cot} \theta =\frac{\sqrt{5}}{2}$ seen as the final answer
         The R1 should be awarded independently for a negative value only given as a final answer.

 

METHOD 2

Attempt to use $1+{\text{cot}}^2 \theta ={\text{cosec}}^2 \theta$        M1

$1+{\text{cot}}^2 \theta =\frac{9}{4}$

$\text{cot}{ }^2\theta =\frac{5}{4}$        (A1)

$\text{cot} \theta =\pm \frac{\sqrt{5}}{2}$

$\text{cot} \theta <0$ (since $\text{cosec} \theta >0$ puts $\theta$ in the second quadrant)        R1

$\text{cot} \theta =-\frac{\sqrt{5}}{2}$        A1

Note: Award M1A1R0A0 for $\text{cot} \theta =\frac{\sqrt{5}}{2}$ seen as the final answer
         The R1 should be awarded independently for a negative value only given as a final answer.

 

METHOD 3

$\text{sin} \theta =\frac{2}{3}$

attempt to use ${\text{sin}}^2 \theta +{\text{cos}}^2 \theta =1$        M1

$\frac{4}{9}+{\text{cos}}^2 \theta =1$

${\text{cos}}^2 \theta =\frac{5}{9}$        (A1)

$\text{cos} \theta =\pm \frac{\sqrt{5}}{3}$

$\text{cos} \theta <0$ (since $\text{cosec} \theta >0$ puts $\theta$ in the second quadrant)        R1

$\text{cos} \theta =-\frac{\sqrt{5}}{3}$

$\text{cot} \theta =-\frac{\sqrt{5}}{2}$        A1

 

Note: Award M1A1R0A0 for $\text{cot} \theta =\frac{\sqrt{5}}{2}$ seen as the final answer
         The R1 should be awarded independently for a negative value only given as a final answer.

 

[4 marks]

[N/A]