Date | May 2021 | Marks available | 4 | Reference code | 21M.1.AHL.TZ1.6 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 1 |
Command term | Find | Question number | 6 | Adapted from | N/A |
Question
It is given that cosec θ=32, where π2<θ<3π2. Find the exact value of cot θ.
Markscheme
METHOD 1
attempt to use a right angled triangle M1
correct placement of all three values and θ seen in the triangle (A1)
cot θ<0 (since cosec θ>0 puts θ in the second quadrant) R1
cot θ=-√52 A1
Note: Award M1A1R0A0 for cot θ=√52 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+cot2 θ=cosec2 θ M1
1+cot2 θ=94
cot 2θ=54 (A1)
cot θ=±√52
cot θ<0 (since cosec θ>0 puts θ in the second quadrant) R1
cot θ=-√52 A1
Note: Award M1A1R0A0 for cot θ=√52 seen as the final answer
The R1 should be awarded independently for a negative value only given as a final answer.
METHOD 3
sin θ=23
attempt to use sin2 θ+cos2 θ=1 M1
49+cos2 θ=1
cos2 θ=59 (A1)
cos θ=±√53
cos θ<0 (since cosec θ>0 puts θ in the second quadrant) R1
cos θ=-√53
cot θ=-√52 A1
Note: Award M1A1R0A0 for cot θ=√52 seen as the final answer
The R1 should be awarded independently for a negative value only given as a final answer.
[4 marks]