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CBSE · Class 10 · Mathematics

Introduction to Trigonometry 8.4: Trigonometric Identities

5 concepts · 40 practice questions with answers · Free to read

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sin²A + cos²A = 1

From daily life

Square tiles on the two short sides of a right-angled corner equal the tiles on the long side. Dividing by the long side’s tiles gives this identity.

Simple definition

An identity is true for every value of the angle. Dividing a² + b² = c² by c² gives sin²A + cos²A = 1. So sin²A = 1 − cos²A and cos²A = 1 − sin²A.

Formula: sin²A + cos²A = 1

ABCbaca² + b² = c²÷ c²(a/c)² + (b/c)² = 1

In the figure: Divide a² + b² = c² by c²: (a/c)² + (b/c)² = 1.

Live figure

sin²A + cos²A = 1

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Rule 1: Check the identity — Put in known values and add the squares.

  • Easy: sin A = 3/5, cos A = 4/5: 9/25 + 16/25 = 1.
  • Medium: At 60°: (√3/2)² + (1/2)² = 3/4 + 1/4 = 1.
  • Exam-type: At 0°: 0² + 1² = 1.

Rule 2: Find one from the other — Square the known ratio, subtract from 1, take the positive root.

  • Easy: sin A = 5/13: cos²A = 1 − 25/169 = 144/169, so cos A = 12/13.
  • Medium: cos A = 3/5: sin²A = 1 − 9/25 = 16/25, so sin A = 4/5.
  • Exam-type: For an acute angle take only the positive root.
Remember

The identity holds for every angle, so it is stronger than checking a few values.

Common mistake

Writing sin²A as sin A²; it means (sin A)², the square of the ratio.

Memory clue

"Sin-squared plus cos-squared is one, always."

Practice questions

Easy
  1. 1Find the value of 9 − sin² 70° − cos² 70°.
  2. 2Find the value of 3 − sin² θ − cos² θ.
  3. 3If cos θ = 8/17 for an acute angle θ, find sin θ.
Medium
  1. 4If cos θ = 5/13 for an acute angle θ, find sin θ.
  2. 5If 1 − cos² θ = 64/289, find the value of cos² θ.
  3. 6If cos² θ = 225/289, find sin² θ.
Challenging
  1. 7If cos θ = 8/17 for an acute angle θ, find the value of 1 − 2 sin² θ.
  2. 8If sin θ = 3/5 for an acute angle θ, find the value of 1 − 2 sin² θ.
Show answers (try them yourself first)
  1. 18
  2. 22
  3. 315/17
  4. 412/13
  5. 5225/289
  6. 664/289
  7. 7−161/289
  8. 87/25

1 + tan²A = sec²A and 1 + cot²A = cosec²A

From daily life

Dividing the same Pythagoras sentence by different sides gives different, equally useful, forms.

Simple definition

Divide sin²A + cos²A = 1 by cos²A to get 1 + tan²A = sec²A. Divide it by sin²A to get 1 + cot²A = cosec²A. These hold wherever the ratios are defined.

Formula: 1 + tan²A = sec²A; 1 + cot²A = cosec²A

sin²A + cos²A = 11 + tan²A = sec²A1 + cot²A = cosec²A

In the figure: Three identities to remember.

Rule 1: Use 1 + tan²A = sec²A — Square tan, add 1, take the root for sec.

  • Easy: tan A = 3/4: sec²A = 1 + 9/16 = 25/16, so sec A = 5/4.
  • Medium: At 45°: 1 + 1 = 2 = (√2)².
  • Exam-type: sec²A − tan²A = 1 is the same identity rearranged.

Rule 2: Use 1 + cot²A = cosec²A — Square cot, add 1, take the root for cosec.

  • Easy: cot A = 4/3: cosec²A = 1 + 16/9 = 25/9, so cosec A = 5/3.
  • Medium: At 30°: 1 + 3 = 4 = 2².
  • Exam-type: cosec²A − cot²A = 1 is the rearranged form.
Remember

The right pairs: tan with sec, cot with cosec.

Common mistake

Pairing tan with cosec, or forgetting the 1 in 1 + tan²A.

Memory clue

"One plus tan-square is sec-square; one plus cot-square is cosec-square."

Practice questions

Easy
  1. 1If cot θ = 24/7 for an acute angle θ, find cosec θ.
  2. 2The identity 1 + cot² A = ___
  3. 3If tan θ = 8/15 for an acute angle θ, find sec θ.
Medium
  1. 4If tan θ = 5/12 for an acute angle θ, find cos θ.
  2. 5If cosec θ = 13/5 for an acute angle θ, find sin θ.
  3. 6If cot θ = 3/4 for an acute angle θ, find sin θ.
Challenging
  1. 7If tan θ = 21/20 for an acute angle θ, find the value of sec θ + tan θ.
  2. 8If sin θ = 21/29 for an acute angle θ, find the value of sec θ − tan θ.
Show answers (try them yourself first)
  1. 125/7
  2. 2cosec² A
  3. 317/15
  4. 412/13
  5. 55/13
  6. 64/5
  7. 75/2
  8. 82/5

Simplifying expressions

From daily life

A long phone number is easier to remember once grouped. Identities group long trigonometric expressions into short ones.

Simple definition

To simplify, convert everything to sin and cos, or spot a pattern like 1 − sin²A and replace it by an identity. Cancel common factors at the end. To prove an identity, work on one side until it equals the other.

write insin and cosuseidentitycancel

In the figure: (1 + tan²A) cos²A = sec²A × cos²A = 1.

Rule 1: Replace by an identity — Look for 1 − sin²A, 1 − cos²A, 1 + tan²A, 1 + cot²A.

  • Easy: (1 + tan²A) cos²A = sec²A cos²A = 1.
  • Medium: (1 − cos²A) cosec²A = sin²A cosec²A = 1.
  • Exam-type: (1 − sin A)(1 + sin A) = 1 − sin²A = cos²A.

Rule 2: Expand and prove — Expand brackets, group sin² and cos², then use the identity.

  • Easy: (sin A + cos A)² + (sin A − cos A)² = 2 sin²A + 2 cos²A = 2.
  • Medium: (1 + tan²A) ÷ (1 + cot²A) = sec²A ÷ cosec²A = sin²A ÷ cos²A = tan²A.
  • Exam-type: tan A cot A = 1, so tan A cot A + sin²A + cos²A = 2.
Remember

When stuck, rewrite every ratio in sin and cos; the identity usually shows itself.

Common mistake

Cancelling terms across a plus sign; only common factors can be cancelled.

Memory clue

"Everything to sin and cos, then spot the pattern."

Practice questions

Easy
  1. 1Simplify: sec² A − tan² A
  2. 2Simplify: (1 − cos A)(1 + cos A)
  3. 3Simplify: (1 − cos B)(1 + cos B)
Medium
  1. 4Simplify: cosec² B − cot² B
  2. 5Simplify: (cosec x − 1)(cosec x + 1)
  3. 6Simplify: (1 + cot² α) sin² α
Challenging
  1. 7Simplify: cosec² θ − cot² θ
  2. 8Simplify: sin α sec α cot α
Show answers (try them yourself first)
  1. 11
  2. 2sin² A
  3. 3sin² B
  4. 41
  5. 5cot² x
  6. 61
  7. 71
  8. 81

Expressing one ratio in terms of another

From daily life

Converting rupees to dollars needs one rate. Here the identities are the “rates” that convert one ratio into another.

Simple definition

To write every ratio in terms of one, say sin A, first find cos A = √(1 − sin²A), then build tan A = sin A ÷ cos A and the reciprocals. For any other starting ratio, use the matching identity.

Formula: cos A = √(1 − sin²A); tan A = sin A/√(1 − sin²A); sec A = √(1 + tan²A)

start: sin Acos A = √(1 − sin²A)tan A = sin A ÷ cos A

In the figure: Start with sin A, find cos A, then tan A and the rest.

Rule 1: In terms of sin A — Use cos A = √(1 − sin²A) and divide.

  • Easy: cot A = √(1 − sin²A) ÷ sin A.
  • Medium: sec A = 1 ÷ √(1 − sin²A).
  • Exam-type: Check with sin A = 4/5: cos A = 3/5 and tan A = 4/3.

Rule 2: In terms of tan A or cot A — Use 1 + tan²A = sec²A, then take reciprocals.

  • Easy: sec A = √(1 + tan²A) and cos A = 1 ÷ √(1 + tan²A).
  • Medium: sec A in terms of cot A: √(cot²A + 1) ÷ cot A.
  • Exam-type: Check with cot A = 3/4: sec A = (5/4) ÷ (3/4) = 5/3.
Remember

A square root appears; for an acute angle always take the positive value.

Common mistake

Leaving out the square root, e.g. writing cos A = 1 − sin²A instead of √(1 − sin²A).

Memory clue

"Square, subtract or add one, root: the identity gives the next ratio."

Practice questions

Easy
  1. 1If sec θ = 5/4 for an acute angle θ, use the identities to find sin θ.
  2. 2If cos θ = 5/13 for an acute angle θ, use the identities to find cosec θ.
  3. 3If cot θ = 8/15 for an acute angle θ, use the identities to find cos θ.
Medium
  1. 4Express cot α in terms of cos α (α is acute).
  2. 5Express cos α in terms of cot α (α is acute).
  3. 6Express sin θ in terms of cos θ (θ is acute).
Challenging
  1. 7If cos θ = 8/17 for an acute angle θ, find the value of sec θ cosec θ.
  2. 8If cos θ = 4/5 for an acute angle θ, find the value of (1 − sin θ)/(1 + sin θ).
Show answers (try them yourself first)
  1. 13/5
  2. 213/12
  3. 38/17
  4. 4cos α / √(1 − cos² α)
  5. 5cot α / √(1 + cot² α)
  6. 6√(1 − cos² θ)
  7. 7289/120
  8. 81/4

Using identities in problems

From daily life

A detective combines two clues to reach a third. Two ratio facts can be combined with an identity to find the missing value.

Simple definition

Use the identity to connect the given ratio to the one asked for. A difference of squares such as sec²A − tan²A = 1 factors as (sec A − tan A)(sec A + tan A) = 1, which gives one value from the other.

sec²A − tan²A = 1(sec A − tan A)(sec A + tan A) = 1sec A + tan A = 2 → sec A − tan A = 1/2sec A = 5/4, tan A = 3/4

In the figure: If sec A + tan A = 2 then sec A − tan A = 1/2, so sec A = 5/4 and tan A = 3/4.

Rule 1: Find a value — Find the missing ratios first, then evaluate.

  • Easy: If sin A = 3/5, then cos A = 4/5, so sec A + tan A = 5/4 + 3/4 = 2.
  • Medium: 9 sec²A − 9 tan²A = 9 × 1 = 9.
  • Exam-type: (cosec A − cot A)(cosec A + cot A) = cosec²A − cot²A = 1.

Rule 2: Use the product trick — Multiply the given sum by the unknown difference.

  • Easy: sec A + tan A = 2 gives sec A − tan A = 1/2.
  • Medium: Adding: 2 sec A = 5/2, so sec A = 5/4; subtracting gives tan A = 3/4.
  • Exam-type: Check: (5/4)² − (3/4)² = 25/16 − 9/16 = 1.
Remember

Whenever a squared ratio appears with a 1, think of the three identities first.

Common mistake

Treating (sec A + tan A)² as sec²A + tan²A; the middle term is missing in that.

Memory clue

"Square-difference equals one: sum times difference is one."

Practice questions

Easy
  1. 1Find the value of 5 − (sec² A − tan² A).
  2. 2Find the value of 5 − (cosec² B − cot² B).
  3. 3Find the value of 4 (cosec² θ − cot² θ).
Medium
  1. 4Find the value of sec² B − tan² B + 5.
  2. 5If tan θ + cot θ = 25/12, find the value of (tan θ − cot θ)².
  3. 6If tan θ + cot θ = 61/30, find the value of (tan θ − cot θ)².
Challenging
  1. 7If cosec θ − cot θ = 3/5, find cosec θ + cot θ.
  2. 8If sin θ + cos θ = 41/29 (θ acute), find sin θ cos θ.
Show answers (try them yourself first)
  1. 14
  2. 24
  3. 34
  4. 46
  5. 549/144
  6. 6121/900
  7. 75/3
  8. 8420/841