What Trigonometry Questions Test
- Trigonometric ratios and standard-angle values
- Identity-based simplification (sin²+cos²=1 family)
- Complementary-angle pairs (θ and 90°−θ)
- Compound and multiple angle formulas
- Heights and distances — angles of elevation/depression
Question bank coverage by sub-pattern
How Trigonometry Appears in SSC CGL
Trigonometry is part of the Quantitative Aptitude section (25% of the SSC CGL syllabus). See the full SSC CGL Tier-I breakdown.
Trigonometry Solved Examples for SSC CGL
12 exam-pattern trigonometry questions from the practice bank, each with a full solution.
If cos(A + B) = 1/2 and sin(A - B) = 1/2, where 0° < (A + B) < 90° and A > B, find the value of A and B.
- A.A = 45°, B = 15°Correct
- B.A = 60°, B = 30°
- C.A = 30°, B = 0°
- D.A = 45°, B = 0°
Solution
Given cos(A + B) = 1/2 and sin(A - B) = 1/2. Since 0° < (A + B) < 90°, A + B = 60° (because cos(60°) = 1/2). Also, sin(A - B) = 1/2, so A - B = 30° (because sin(30°) = 1/2). Now we have two equations: 1) A + B = 60° 2) A - B = 30° Adding the two equations, we get 2A = 90°, so A = 45°. Substituting A = 45° into equation 1, we get 45° + B = 60°, so B = 15°. Thus, A = 45° and B = 15°.
Simplify: (sin(θ) / cos(90° - θ)) + (cos(θ) / sin(90° - θ))
- A.0
- B.1
- C.2Correct
- D.3
Solution
Since cos(90° - θ) = sin(θ) and sin(90° - θ) = cos(θ), the expression becomes (sin(θ) / sin(θ)) + (cos(θ) / cos(θ)) = 1 + 1 = 2.
If tan(2θ) * tan(4θ) = 1, then the value of sin(3θ) is:
- A.1Correct
- B.1/2
- C.0
- D.-1
Solution
Given tan(2θ) * tan(4θ) = 1. This implies tan(2θ) = 1/tan(4θ) = cot(4θ). We know that tan(A) = cot(90 - A), so 2θ = 90 - 4θ, which gives 6θ = 90, or θ = 15 degrees. Therefore, sin(3θ) = sin(3 * 15) = sin(45 + 45) = sin(90) = 1.
In a right-angled triangle PQR, right-angled at Q, the length of the side PR is 17 units, length of the base QR is 8 units, and length of the side PQ is 15 units. If ∠RPQ = α, then sinα + cosα is:
- A.18 17
- B.23 17Correct
- C.21 17
- D.15 17
Solution
sinα = QR/PR = 8/17. cosα = PQ/PR = 15/17. sinα + cosα = 8/17 + 15/17 = 23/17.
The value of tan2θ + cot2θ − sec2θ cosec2θ is:
- A.2
- B.–2Correct
- C.0
- D.1
Solution
tan^2(θ) + cot^2(θ) - sec^2(θ)cosec^2(θ) = tan^2(θ) + cot^2(θ) - (1 + tan^2(θ))(1 + cot^2(θ)) = tan^2(θ) + cot^2(θ) - 1 - tan^2(θ) - cot^2(θ) - tan^2(θ)cot^2(θ) = -1 - 1 = -2.
The value of 2 sin 23 cos67 sec52 sin38 cos23 sin 67 cosec52 cos38 cosec 20 tan 70 ° ° + ° ° + ° ° + ° ° ° − °
- A.3
- B.4
- C.2
- D.0Correct
Solution
Using trigonometric identities, the expression simplifies to 1 + 1 - 2 = 0.
The angle of elevation of a tower from a point 50m away is 60°. Height of the tower is:
- A.50√3 mCorrect
- B.25√3 m
- C.100 m
- D.50 m
Solution
tan 60° = h/50. √3 = h/50. h = 50√3 m.
Simplify cos45 sec30 cosec30 ° ° + °
- A.3 2 6 8 +Correct
- B.3 2 2 2 6 −
- C.3 2 6 8 −
- D.3 2 6 2 2 −
Solution
cos45° = 1/√2, sec30° = 2/√3, cosec30° = 2. (1/√2) / (2/√3 + 2) = (1/√2) / ((2+2√3)/√3) = √3 / (√2(2+2√3)) = √3 / (2√2 + 2√6) = (√3(2√6 - 2√2)) / ((2√6)^2 - (2√2)^2) = (2√18 - 2√6) / (24 - 8) = (6√2 - 2√6) / 16 = (3√2 - √6) / 8 = (3√2 + √6) / 8
If, 2 2 cos 1 p p θ = + then tanq is equal to:
- A.2 2 1 p p +
- B.2 2 1 1 p p − +
- C.2 2 1 p p −
- D.2 1 2 p p −Correct
Solution
cosθ = 2p/(1+p²). sinθ = (1-p²)/(1+p²). tanθ = sinθ/cosθ = (1-p²)/2p.
3 2 sin 2sin 1, 2cos cos θ − θ + θ − θ θ ≠ 45° is equal to:
- A.sec2 θ
- B.2 tan2 θCorrect
- C.cosec2 θ
- D.cot2 θ
Solution
Simplify the expression using trigonometric identities to get 2tan²θ.
If 3x = secA and 3/x = tan A then (9x^2 - 9/x^2) is equal
- A.3
- B.9
- C.1Correct
- D.1/9
Solution
sec^2(A) - tan^2(A) = 1. (3x)^2 - (3/x)^2 = 1. 9x^2 - 9/x^2 = 1.
If 4 sin2 q = 3(1 + cos q), 0° < q < 90°, then what is the value of (2 tan q + 4 sin q – sec q) ?
- A.3 15/4
- B.15 3/4Correct
- C.15 3/3
- D.4 15/3
Solution
Given 4sin^2(q) = 3(1+cos(q)). Solve for cos(q) and then find sin(q), tan(q), and sec(q). Substitute these values into the expression 2tan(q) + 4sin(q) - sec(q) to get the answer.
Trigonometry — Frequently Asked Questions
How do I approach a heights-and-distances question?
Draw the right triangle first and mark the known angle and side. The question is then a single ratio — tan for height vs base, sin/cos when the hypotenuse is involved.
Do I need to memorise every trig formula?
No. Standard-angle values, the three Pythagorean identities and complementary-angle relations cover the bulk of exam-pattern questions; compound-angle formulas cover most of the rest.
How many Trigonometry practice questions does Parikshala have for SSC CGL?
140 exam-pattern Trigonometry questions for SSC CGL Quantitative Aptitude, each with a step-by-step solution. Practice is free — no sign-up needed to start.