IB · Mathematics: Analysis & Approaches SL & HL · Geometry & trigonometry

Compound angles & reciprocal trig: IB Mathematics: Analysis & Approaches questions with answers

Free Compound angles & reciprocal trig questions for IB Mathematics: Analysis & Approaches (SL & HL), HL only, part of Geometry & trigonometry. Each question has the answer and the marking points an examiner would look for.

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Compound angles & reciprocal trig practice questions

Question 1Structured · 6 marksA* challenge

Solve the equation 2cot2x + 3 = 3cosec x for 0 ≤ x < 2π.

[6]
Show answers and mark scheme
PartAnswer and marking pointsMarks
x = π2 (cosec x = 12 is impossible)
use of a Pythagorean identity (1 + tan2x = sec2x or 1 + cot2x = cosec2x) M1
2cosec2x − 3cosec x + 1 = 0 A1
(2cosec x − 1)(cosec x − 1) = 0 A1
attempt to find all solutions in the interval (M1)
x = π2 (cosec x = 12 is impossible) A2
6
Question 2Structured · 5 marks

The angles A and B are acute, with tan A = 23 and tan B = 18.

(a)

Find the exact value of tan(A + B).

[3]
(b)

Find the exact value of cos A.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)1922
tan A + tan B1 − tan A tan B (M1)
19/2411/12 (A1)
1922 A1
3
(b)3√13
right-angled triangle with opposite 2, adjacent 3 (M1)
cos A = 3√13 A1
2
Question 3Structured · 6 marksA* challenge

Solve the equation 2sec2x − 3tan x = 1 for 0 ≤ x < 2π.

[6]
Show answers and mark scheme
PartAnswer and marking pointsMarks
x = π4, 5π4, arctan12, π + arctan12
use of a Pythagorean identity (1 + tan2x = sec2x or 1 + cot2x = cosec2x) M1
2tan2x − 3tan x + 1 = 0 A1
(2tan x − 1)(tan x − 1) = 0 A1
attempt to find all solutions in the interval (M1)
x = π4, 5π4, arctan12, π + arctan12 A2
6
Question 4Structured · 5 marks

The angles A and B are acute, with tan A = 12 and tan B = 15.

(a)

Find the exact value of tan(A + B).

[3]
(b)

Find the exact value of cos A.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)79
tan A + tan B1 − tan A tan B (M1)
7/109/10 (A1)
79 A1
3
(b)2√5
right-angled triangle with opposite 1, adjacent 2 (M1)
cos A = 2√5 A1
2
Question 5Structured · 6 marksA* challenge

Solve the equation cosec2x = 3cot x − 1 for 0 ≤ x < 2π.

[6]
Show answers and mark scheme
PartAnswer and marking pointsMarks
x = π4, 5π4, arctan12, π + arctan12
use of a Pythagorean identity (1 + tan2x = sec2x or 1 + cot2x = cosec2x) M1
cot2x − 3cot x + 2 = 0 A1
(cot x − 1)(cot x − 2) = 0 A1
attempt to find all solutions in the interval (M1)
x = π4, 5π4, arctan12, π + arctan12 A2
6
Question 6Structured · 5 marks

The angles A and B are acute, with tan A = 23 and tan B = 17.

(a)

Find the exact value of tan(A + B).

[3]
(b)

Find the exact value of cos A.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)1719
tan A + tan B1 − tan A tan B (M1)
17/2119/21 (A1)
1719 A1
3
(b)3√13
right-angled triangle with opposite 2, adjacent 3 (M1)
cos A = 3√13 A1
2

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in