Trigonometry
🔒 Log in to trackRatios, the standard-value table, the three identities, complementary angles, value-putting and max-min values. Tier 1 reliably holds 2-4 trig questions and they fall almost mechanically to the value table, one identity, or a 3-4-5 style triangle — among the cheapest marks in the paper.
One page per subtopic: detailed notes, every question type, formulas, tricks and practice sets.
Every formula on one printable page, grouped by subtopic.
5 exam-level questions worked step by step.
72 questions — untimed practice or a timed test with analysis.
Track record in the exam
Questions per shift in recent SSC CGL papers.
Test difficulty mix (72 questions)
Question patterns exams keep repeating
Taken from previous-year papers. If a pattern is marked "very common", expect to see it in your exam.
Values at 30°, 45° and 60°
very commonA sum or product of sin, cos, tan at 0°, 30°, 45°, 60° or 90°.
Learn the small table: sin 30° = ½, cos 60° = ½, tan 45° = 1.
Put in the values and work out the sum.
Find sin 30° + cos 60° + tan 45°.
sin 30° = ½
cos 60° = ½
tan 45° = 1
Sum = ½ + ½ + 1 = 2
One ratio is given, find another
very common"If sin θ = 3/5, find tan θ."
Draw a right triangle and write the given sides on it.
Find the third side with Pythagoras, then read off the ratio.
If sin θ = 3/5, find tan θ.
sin θ = opposite ÷ hypotenuse, so opposite = 3, hypotenuse = 5
Third side = √(25 − 9) = 4 (adjacent)
tan θ = opposite ÷ adjacent = 3/4
sec + tan (or cosec + cot) is given
very common"If sec θ + tan θ = 3, find sec θ − tan θ."
The two always multiply to 1.
So the other one is 1 ÷ the given value.
If sec θ + tan θ = 3, find sec θ − tan θ.
(sec θ + tan θ) × (sec θ − tan θ) = 1
3 × (sec θ − tan θ) = 1
sec θ − tan θ = 1/3
sin θ + cos θ is given
very common"If sin θ + cos θ = 7/5, find sin θ cos θ."
Square both sides.
Use sin² θ + cos² θ = 1 to get the product.
If sin θ + cos θ = 7/5, find sin θ cos θ.
Square: sin² + cos² + 2 sin cos = 49/25
sin² + cos² = 1, so 2 sin cos = 49/25 − 1 = 24/25
sin cos = 24/25 ÷ 2 = 12/25
tan θ is given, find a sin–cos fraction
very common"If tan θ = 2, find (3 sin θ + cos θ)/(sin θ + cos θ)."
Divide the top and bottom by cos θ.
Every sin θ ÷ cos θ becomes tan θ. Then put in the number.
If tan θ = 2, find (3 sin θ + cos θ) ÷ (sin θ + cos θ).
Divide top and bottom by cos θ:
(3 tan θ + 1) ÷ (tan θ + 1)
Put tan θ = 2: (3 × 2 + 1) ÷ (2 + 1)
= 7 ÷ 3 = 7/3
Angles that add up to 90°
very commonA long product like tan 10° × tan 20° × tan 70° × tan 80°.
Pair angles that add up to 90°.
Each pair multiplies to 1.
Find tan 10° × tan 20° × tan 70° × tan 80°.
10° + 80° = 90°, so tan 10° × tan 80° = 1
20° + 70° = 90°, so tan 20° × tan 70° = 1
Product = 1 × 1 = 1
Find the angle from an equation like tan = cot
very common"If tan 2θ = cot(θ − 12°), find θ."
tan and cot of two angles are equal only when the angles add up to 90°.
Solve the simple equation.
If tan 2θ = cot(θ − 12°), find θ.
The two angles add up to 90°
2θ + θ − 12° = 90°
3θ = 102°
θ = 34°
Biggest value of a sin θ + b cos θ
very common"Find the maximum value of 8 sin θ − 15 cos θ."
Maximum = √(a² + b²).
Minimum is the same number with a minus sign.
Find the maximum value of 8 sin θ − 15 cos θ.
8² = 64 and 15² = 225
64 + 225 = 289
√289 = 17, so the maximum is 17
(the minimum is −17)
Smallest value of tan θ + cot θ
common"Find the minimum value of tan θ + cot θ" (or tan² + cot²).
A number plus its reciprocal is never below 2.
It equals 2 when the number is 1 (θ = 45°).
Find the minimum value of tan θ + cot θ for an acute angle θ.
At 45°: tan = 1, cot = 1, sum = 2
At 30°: 1/√3 + √3 is about 2.31, which is bigger
A number plus its reciprocal is never below 2
Minimum = 2
Two sides of a right triangle are given
commonTwo sides of a right triangle are given. Asks the third side or a ratio.
Find the third side with Pythagoras.
The 8-15-17 and 3-4-5 triples often appear.
A right triangle has legs 15 cm and 8 cm. Find the hypotenuse.
15² = 225 and 8² = 64
225 + 64 = 289
√289 = 17 cm