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State Board · Class 11 · Mathematics Part 1

Lesson 3: Trigonometry - II

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

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Sum and difference of two angles (compound angles)

From daily life

Lift a shop shutter by 30° and then by 45° more and the total lift is 75°. But sin 75° is not simply sin 30° + sin 45°; a special formula is needed. Building on classes 9–10 : In class 10 we knew only 30°, 45°, 60°; now these formulas give 15°, 75°, 105° and similar angles.

Simple definition

If A and B are two angles, then A + B and A − B are called compound angles. The formulas below give their sin, cos, tan in terms of the functions of A and B.

Rule 1: Formulas for sin and cos — sin(A + B) = sin A cos B + cos A sin B, sin(A − B) = sin A cos B − cos A sin B, cos(A + B) = cos A cos B − sin A sin B, cos(A − B) = cos A cos B + sin A sin B.

Rule 2: Formula for tan — tan(A + B) = (tan A + tan B)/(1 − tan A tan B) and tan(A − B) = (tan A − tan B)/(1 + tan A tan B).

XYP(cos A, sin A)Q(cos B, sin B)BA − BAO

In the figure : On the unit circle P(cos A, sin A) and Q(cos B, sin B); the angle between OP and OQ is A − B.

How is the formula obtained? : On the unit circle take P(cos A, sin A) and Q(cos B, sin B). By the distance formula PQ² = (cos A − cos B)² + (sin A − sin B)² = 2 − 2(cos A cos B + sin A sin B). Now treat OQ as a new X-axis: P has co-ordinates (cos(A − B), sin(A − B)) and Q has (1, 0), so PQ² = 2 − 2 cos(A − B). Equating the two values of PQ² gives cos(A − B) = cos A cos B + sin A sin B. The rest follow: replace B by −B for cos(A + B); use 90° − A for sin(A ± B); divide for tan.

  • Easy : sin 15° = sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30° = (1/√2)(√3/2) − (1/√2)(1/2) = (√6 − √2)/4.
  • Medium : sin A = 8/17, cos B = 3/5 (both acute): cos A = 15/17, sin B = 4/5; sin(A + B) = (8/17)(3/5) + (15/17)(4/5) = 24/85 + 60/85 = 84/85.
  • Exam-type : If sin A = 4/5 and sin B = 3/5 (acute), find sin(A + B) sin(A − B): it equals sin²A − sin²B = 16/25 − 9/25 = 7/25.
Remember

In the cos formulas the sign between the products is opposite to the sign in the angle; in sin it is the same. In tan the numerator has the angle's sign and the denominator the opposite.

Common mistake

Writing sin(A + B) = sin A + sin B. sin is not a multiplier; sin(A + B) always expands into two products of four functions.

Shortcut

sin: "sin-cos + cos-sin" (same sign). cos: "cos-cos − sin-sin" (opposite sign). The pairs always mix the sin of one angle with the cos of the other. Memory clue : "sin-cos, cos-sin; cos-cos, sin-sin — cos flips the sign."

CET trap

CET gives tan A and tan B and asks for A + B; if tan(A + B) = 1 then A + B = 45° for acute angles, but it could be 225° if the angles are in other quadrants; check the stated range.

CET-style example : If tan A = 1/2 and tan B = 1/3 (A, B acute), then A + B = ? (A) 30° (B) 45° (C) 60° (D) 90°

Answer : (B). tan(A + B) = (1/2 + 1/3)/(1 − 1/6) = (5/6)/(5/6) = 1, so A + B = 45°. Time-saving method : When numerator and denominator come out equal (5/6) the value is 1; recognise tan 45° = 1 at once.

Spot the difference

Anglesincostan
A + Bsin A cos B + cos A sin Bcos A cos B − sin A sin B(tan A + tan B)/(1 − tan A tan B)
A − Bsin A cos B − cos A sin Bcos A cos B + sin A sin B(tan A − tan B)/(1 + tan A tan B)

Practice questions

Easy
  1. 1Find the value of cos 15°.
  2. 2Find the value of sin 40° cos 20° + cos 40° sin 20°.
  3. 3Write the formula for tan(A − B).
Medium
  1. 4If sin A = 5/13 and cos B = 4/5 (acute), find cos(A − B).
  2. 5If tan A = 3 and tan B = 2, find tan(A − B).
  3. 6Find the value of cos 105°.
Challenging
  1. 7Prove: tan 20° + tan 25° + tan 20° tan 25° = 1.
  2. 8If cos A = 3/5 and sin B = 2/3, find cos(A + B) cos(A − B).
Show answers (try them yourself first)
  1. 1(√6 + √2)/4.
  2. 2√3/2 (this is sin 60°).
  3. 3(tan A − tan B)/(1 + tan A tan B).
  4. 463/65.
  5. 51/7.
  6. 6(√2 − √6)/4.
  7. 7tan 45° = 1 and tan(20° + 25°) = (tan 20° + tan 25°)/(1 − tan 20° tan 25°); cross-multiply to prove.
  8. 8−19/225.

Allied angles

From daily life

Look at the same point of a wheel in a mirror and left and right swap while the height stays the same. Angles near 90°, 180°, 270°, 360° are such mirror images. Building on classes 9–10 : In the last lesson we used the reference angle; now we turn that idea into a compact rule.

Simple definition

Allied angles are angles such as 90° ± θ, 180° ± θ, 270° ± θ, 360° − θ; their functions can be written in terms of the functions of θ.

Rule 1: Rule for name and sign — Treat θ as acute. For 90° or 270° the name of the function changes (sin↔cos, tan↔cot, sec↔cosec). For 180° or 360° the name stays. The sign is the sign of the original function in the resulting quadrant.

XYP₁ (cos θ, sin θ)θ(−cos θ, sin θ)180° − θ(−cos θ, −sin θ)180° + θ(cos θ, −sin θ)360° − θO

In the figure : The points for θ, 180° − θ, 180° + θ, 360° − θ are mirror images of one another; the signs of the co-ordinates give the signs of sin and cos.

How is the formula obtained? : Let P(x, y) be the point for θ. Its reflection in the Y-axis, (−x, y), is the point for 180° − θ, so sin(180° − θ) = y = sin θ and cos(180° − θ) = −x = −cos θ. Its reflection in the line y = x, (y, x), is the point for 90° − θ; x and y swap, so sin(90° − θ) = cos θ and cos(90° − θ) = sin θ.

Anglesincostan
90° − θcos θsin θcot θ
90° + θcos θ− sin θ− cot θ
180° − θsin θ− cos θ− tan θ
180° + θ− sin θ− cos θtan θ
270° − θ− cos θ− sin θcot θ
270° + θ− cos θsin θ− cot θ
360° − θ− sin θcos θ− tan θ
  • Easy : sin(90° + θ) = cos θ: 90° + θ is in quadrant II, sin is positive and the name becomes cos.
  • Medium : cos(270° + θ) = sin θ: 270° + θ is in quadrant IV, cos is positive; 270° changes the name to sin.
  • Exam-type : cos(90° + θ) sin(180° − θ) / [tan(180° + θ) cos(270° − θ)]: numerator = (−sin θ)(sin θ), denominator = (tan θ)(−sin θ); the quotient is sin θ/tan θ = cos θ.
Remember

180° and 360°: name stays. 90° and 270°: name changes. The sign is always that of the original function in the quadrant obtained by treating θ as acute.

Common mistake

Deciding the sign from the new function. In sin(90° + θ) = cos θ the sign comes from sin (positive in quadrant II), not from cos.

Shortcut

Count the multiple of 90° in the angle: odd (90°, 270°) → change the name; even (180°, 360°) → keep the name. Then put the sign from the quadrant. Memory clue : "Odd swaps, even stays."

CET trap

CET uses large angles like 900° and −690°; first reduce by 360°, then apply the rule. For tan a remainder with 180° also works.

CET-style example : What is the value of sin 600° cos 330° + cos 600° sin 330°? (A) −1/2 (B) 1/2 (C) √3/2 (D) 1

Answer : (A). It is sin(600° + 330°) = sin 930° = sin(930° − 720°) = sin 210° = −1/2. Time-saving method : Recognise the pattern sin A cos B + cos A sin B; make one sin(A + B) and reduce the angle once.

Spot the difference

AngleNameSign from
90° ± θchangesoriginal function, quadrant II / I
180° ± θsameoriginal function, quadrant II / III
270° ± θchangesoriginal function, quadrant III / IV
360° − θsameoriginal function, quadrant IV

Practice questions

Easy
  1. 1What is sin(180° + θ) equal to?
  2. 2Find the value of cos 150°.
  3. 3What is tan(90° − θ) equal to?
Medium
  1. 4Find the value of sin 240°.
  2. 5Simplify cos(270° − θ) + sin(180° + θ).
  3. 6Find the value of tan 315°.
Challenging
  1. 7Find the value of sin 780° cos 390°.
  2. 8Simplify sec(90° + θ) cosec(180° − θ).
Show answers (try them yourself first)
  1. 1−sin θ.
  2. 2−√3/2.
  3. 3cot θ.
  4. 4−√3/2.
  5. 5−2 sin θ.
  6. 6−1.
  7. 73/4 (sin 60° cos 30°).
  8. 8−cosec²θ (since sec(90° + θ) = −cosec θ and cosec(180° − θ) = cosec θ).

Double, triple and half angles

From daily life

Turn a bicycle pedal twice as far and the wheel turns through twice the angle. Even then sin does not simply double; it has its own formula. Building on classes 9–10 : Putting B = A in the compound-angle formulas of Section 1 gives the double-angle formulas.

Simple definition

Multiple angles are 2A, 3A, ... and the half angle is A/2. Their functions can be written in terms of the functions of A.

Rule 1: Double angle — sin 2A = 2 sin A cos A = 2 tan A/(1 + tan²A); cos 2A = cos²A − sin²A = 2 cos²A − 1 = 1 − 2 sin²A = (1 − tan²A)/(1 + tan²A); tan 2A = 2 tan A/(1 − tan²A).

XYθθP(cos 2θ, sin 2θ)OA(1, 0)cos 2θsin 2θ

In the figure : P is the point for 2θ; its co-ordinates are (cos 2θ, sin 2θ). The half angle θ is visible at the centre.

Rule 2: Triple angle and half angle — sin 3A = 3 sin A − 4 sin³A; cos 3A = 4 cos³A − 3 cos A; tan 3A = (3 tan A − tan³A)/(1 − 3 tan²A). Half angle: sin²A = (1 − cos 2A)/2 and cos²A = (1 + cos 2A)/2.

How is the formula obtained? : Put B = A in cos(A + B): cos 2A = cos²A − sin²A. With sin²A + cos²A = 1 this can also be written 2 cos²A − 1 and 1 − 2 sin²A. Triple angle: sin 3A = sin(2A + A) = sin 2A cos A + cos 2A sin A = 2 sin A cos²A + (1 − 2 sin²A) sin A = 3 sin A − 4 sin³A. Half angle: from 1 − 2 sin²A = cos 2A we get sin²A = (1 − cos 2A)/2.

  • Easy : sin θ = 5/13 (θ acute): cos θ = 12/13; sin 2θ = 2 × (5/13) × (12/13) = 120/169.
  • Medium : tan θ = 3/4: cos 2θ = (1 − 9/16)/(1 + 9/16) = (7/16)/(25/16) = 7/25.
  • Exam-type : tan θ = 1/2: tan 3θ = (3 × 1/2 − 1/8)/(1 − 3 × 1/4) = (11/8)/(1/4) = 11/2.
Remember

Three forms of cos 2A: cos²A − sin²A, 2cos²A − 1, 1 − 2sin²A. Pick the one that uses the function you are given. The half-angle formula is cos 2A turned around.

Common mistake

Writing sin 2A = 2 sin A or cos 2A = 2 cos A. The correct form 2 sin A cos A has both functions; doubling the angle does not double the value.

Shortcut

If sin θ = 3/5, doubling the angle of the 3-4-5 triangle gives 7-24-25: sin 2θ = 24/25, cos 2θ = 7/25, tan 2θ = 24/7. Likewise 5-12-13 → 119-120-169. Memory clue : "Double the angle of 3-4-5 → 7-24-25."

CET trap

When tan θ is given use (1 − tan²θ)/(1 + tan²θ) for cos 2θ; do not find sin θ and cos θ separately. The value of tan 2θ is deliberately placed among the options.

CET-style example : If tan θ = 1/3, then cos 2θ = ? (A) 4/5 (B) 3/5 (C) 5/4 (D) 1/2

Answer : (A). cos 2θ = (1 − 1/9)/(1 + 1/9) = (8/9)/(10/9) = 4/5. Time-saving method : If tan θ = a/b then cos 2θ = (b² − a²)/(b² + a²) = (9 − 1)/(9 + 1) = 8/10. Answer in 8 seconds.

Spot the difference

Form of cos 2AWhen to use
cos²A − sin²Awhen both sin and cos are given
2 cos²A − 1when only cos A is given
1 − 2 sin²Awhen only sin A is given
(1 − tan²A)/(1 + tan²A)when only tan A is given

Practice questions

Easy
  1. 1If sin θ = 4/5 (acute), find sin 2θ.
  2. 2Find the value of 1 − 2 sin²15°.
  3. 3If cos θ = 3/5, find cos 2θ.
Medium
  1. 4Find the value of sin²22.5°.
  2. 5If tan θ = 1/4, find tan 2θ.
  3. 6If sin θ = 1/3, find sin 3θ.
Challenging
  1. 7Prove: cos⁴A − sin⁴A = cos 2A.
  2. 8Find the value of sin 3A/sin A + cos 3A/cos A.
Show answers (try them yourself first)
  1. 124/25.
  2. 2√3/2 (this is cos 30°).
  3. 3−7/25.
  4. 4(2 − √2)/4.
  5. 58/15.
  6. 623/27.
  7. 7cos⁴A − sin⁴A = (cos²A − sin²A)(cos²A + sin²A) = cos 2A × 1.
  8. 84 cos 2A (since 3 − 4 sin²A + 4 cos²A − 3 = 4 cos 2A).

Factorization and defactorization formulae

From daily life

Play two notes of slightly different pitch and together they make one "wavering" note. Turning the sum into a product makes such a combined wave easy to describe. Building on classes 9–10 : In classes 9–10 we turned a² − b² into (a + b)(a − b); now we do the same for sin and cos.

Simple definition

Factorization turns sin C ± sin D or cos C ± cos D into a product. Defactorization turns a product such as 2 sin A cos B into a sum or difference.

Rule 1: Sum to product — sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2); sin C − sin D = 2 cos((C + D)/2) sin((C − D)/2); cos C + cos D = 2 cos((C + D)/2) cos((C − D)/2); cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2).

XCDSMOS = 2 cos((C − D)/2)M = (C + D)/2O

In the figure : The sum OS of the two unit vectors OC and OD points along the middle angle and has length 2 cos((C − D)/2); this shows the formula for cos C + cos D.

Rule 2: Product to sum — 2 sin A cos B = sin(A + B) + sin(A − B); 2 cos A sin B = sin(A + B) − sin(A − B); 2 cos A cos B = cos(A + B) + cos(A − B); 2 sin A sin B = cos(A − B) − cos(A + B).

How is the formula obtained? : Adding sin(A + B) = sin A cos B + cos A sin B and sin(A − B) = sin A cos B − cos A sin B gives 2 sin A cos B = sin(A + B) + sin(A − B). Now put A + B = C and A − B = D; then A = (C + D)/2 and B = (C − D)/2, so sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2). The other formulas come in the same way from cos(A ± B).

  • Easy : sin 50° + sin 10° = 2 sin 30° cos 20° = 2 × (1/2) × cos 20° = cos 20°.
  • Medium : cos 70° − cos 10° = −2 sin 40° sin 30° = −2 × sin 40° × (1/2) = −sin 40°.
  • Exam-type : cos 20° cos 40° cos 80°: cos 20° cos 40° = (1/2)(cos 60° + cos 20°); multiply by cos 80°: (1/4)cos 80° + (1/4)(cos 100° + cos 60°) = 1/8 + (1/4)(cos 80° + cos 100°) = 1/8 + 0 = 1/8.
Remember

cos C − cos D carries a minus sign. In product → sum the left side has the factor 2, so divide by 2 to get the product itself.

Common mistake

Writing sin C − sin D as 2 sin cos. The correct form is 2 cos((C + D)/2) sin((C − D)/2) — cos first, sin second. A wrong pairing changes the answer.

Shortcut

Take half the sum and half the difference of the two angles. sin + sin → sin·cos; cos + cos → cos·cos; sin − sin → cos·sin; cos − cos → −sin·sin. Memory clue : "sin-plus = sin·cos; cos-plus = cos·cos; sin-minus = cos·sin; cos-minus = minus sin·sin."

CET trap

In sums of three or four terms (cos 20° + cos 100° + cos 140°) pair first the two angles whose half-sum is a nice value; usually a term cancels.

CET-style example : sin 80° − sin 20° = ? (A) sin 40° (B) cos 40° (C) sin 50° (D) 2 sin 40°

Answer : (A). sin 80° − sin 20° = 2 cos 50° sin 30° = cos 50° = sin 40°. Time-saving method : Half-sum 50°, half-difference 30°: sin 30° = 1/2 cancels the 2; recognise cos 50° = sin 40°.

Spot the difference

Left sideRight side
sin C + sin D2 sin((C + D)/2) cos((C − D)/2)
sin C − sin D2 cos((C + D)/2) sin((C − D)/2)
cos C + cos D2 cos((C + D)/2) cos((C − D)/2)
cos C − cos D−2 sin((C + D)/2) sin((C − D)/2)

Practice questions

Easy
  1. 1Write sin 70° + sin 20° as a product.
  2. 2Write 2 cos 50° cos 10° as a sum.
  3. 3Find the value of cos 75° + cos 15°.
Medium
  1. 4Find the value of sin 65° − sin 25°.
  2. 5Simplify (sin 7A + sin 3A)/(cos 7A + cos 3A).
  3. 6Find the value of 2 sin 75° sin 15°.
Challenging
  1. 7Find the value of sin 40° sin 80° sin 20°.
  2. 8Prove: (sin A + sin 3A)/(cos A + cos 3A) = tan 2A.
Show answers (try them yourself first)
  1. 12 sin 45° cos 25°.
  2. 2cos 60° + cos 40°.
  3. 3√6/2 (2 cos 45° cos 30°).
  4. 42 cos 45° sin 20° ≈ 0.4837.
  5. 5tan 5A.
  6. 61/2.
  7. 7√3/8.
  8. 8Numerator 2 sin 2A cos A, denominator 2 cos 2A cos A; cancel cos A.

Trigonometric functions of the angles of a triangle

From daily life

However you cut a triangular piece, its three angles always add up to 180° (a straight line). This gives a neat relation among the functions of the three angles. Building on classes 9–10 : Since class 9 we know "the angles of a triangle add up to 180°"; now we derive relations among their functions.

Simple definition

In triangle ABC A + B + C = 180° (π), so A + B = 180° − C. With the allied-angle rules this links the functions of all three angles.

Rule 1: Basic relations — sin(A + B) = sin C, cos(A + B) = − cos C, tan(A + B) = − tan C; also sin((A + B)/2) = cos(C/2) and cos((A + B)/2) = sin(C/2).

BACBCABCℓ ∥ BCA + B + C = 180°

In the figure : Draw a line through A parallel to BC; the angles B, A, C then lie along one straight line at A, so A + B + C = 180°.

Rule 2: Important identities — tan A + tan B + tan C = tan A tan B tan C; cot A cot B + cot B cot C + cot C cot A = 1; sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C; cos A + cos B + cos C = 1 + 4 sin(A/2) sin(B/2) sin(C/2); sin A + sin B + sin C = 4 cos(A/2) cos(B/2) cos(C/2).

How is the formula obtained? : Since A + B = 180° − C, tan(A + B) = −tan C. Writing tan(A + B) = (tan A + tan B)/(1 − tan A tan B) gives tan A + tan B = −tan C (1 − tan A tan B) = −tan C + tan A tan B tan C. Moving −tan C to the left, tan A + tan B + tan C = tan A tan B tan C.

  • Easy : If tan A = 1 and tan B = 2 then tan C = −(1 + 2)/(1 − 2) = 3. Check: 1 + 2 + 3 = 6 = 1 × 2 × 3.
  • Medium : A = 90°, B = 60°, C = 30°: sin 2A + sin 2B + sin 2C = 0 + √3/2 + √3/2 = √3 and 4 sin 90° sin 60° sin 30° = 4 × 1 × (√3/2) × (1/2) = √3.
  • Exam-type : A = B = 45°, C = 90°: cos A + cos B + cos C = √2 and 1 + 4 sin 22.5° sin 22.5° sin 45° = 1 + 4(0.1464)(0.7071) = 1.414. Both agree.
Remember

These identities hold only when A + B + C = 180°. First find the third angle C = 180° − (A + B); then pick the formula.

Common mistake

Using these formulas when the angles add to 90° (the two acute angles of a right triangle). There different relations such as tan A tan B = 1 apply.

Shortcut

For tan: sum of the three = product of the three. For cot: sum of the pair-products = 1. If two angles are known, tan of the third = −tan(A + B). Memory clue : "Sum of three = product of three (tan); cot pair-products add to 1."

CET trap

CET gives tan A and tan B and asks about C. Before computing tan C see whether A + B is a nice angle; if it is 45° then C = 135° at once.

CET-style example : In triangle ABC, if tan A = 1/2 and tan B = 1/3, then tan C = ? (A) 1 (B) −1 (C) 5/6 (D) −5/6

Answer : (B). tan(A + B) = 1, so A + B = 45° and C = 135°; tan 135° = −1. Time-saving method : tan C = −tan(A + B) = −1. Five seconds; no need to substitute into the long identity.

Spot the difference

FunctionSumWhen the three angles add to 180°
tantan A + tan B + tan Cequals the product of the three
cotcot A cot B + cot B cot C + cot C cot Aequals 1
sin 2A, sin 2B, sin 2Csin 2A + sin 2B + sin 2C4 sin A sin B sin C
coscos A + cos B + cos C1 + 4 sin(A/2) sin(B/2) sin(C/2)

Practice questions

Easy
  1. 1In triangle ABC, if A = 50° and B = 60°, find C.
  2. 2What is sin(B + C) equal to?
  3. 3In an equilateral triangle find the value of tan A + tan B + tan C.
Medium
  1. 4If tan A = 2 and tan B = 5, find tan C.
  2. 5For A = B = 45°, C = 90° find sin A + sin B + sin C.
  3. 6For A = 120°, B = C = 30° find cos A + cos B + cos C.
Challenging
  1. 7If A = 60° and tan B + tan C = 4√3, find tan B tan C.
  2. 8Prove: when A + B + C = 180°, cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C.
Show answers (try them yourself first)
  1. 1C = 70°.
  2. 2sin A.
  3. 33√3.
  4. 4tan C = −(2 + 5)/(1 − 10) = 7/9.
  5. 51/√2 + 1/√2 + 1 = 1 + √2.
  6. 6−1/2 + √3/2 + √3/2 = √3 − 1/2.
  7. 7tan B tan C = 5 (√3 + 4√3 = √3 × tan B tan C).
  8. 8cos 2A + cos 2B = −2 cos C cos(A − B) and cos 2C = 2cos²C − 1; use cos C = −cos(A + B).

Lesson 3: Trigonometry - II — quick revision

TopicFormula
Compound anglessin(A ± B) = sin A cos B ± cos A sin B; cos(A ± B) = cos A cos B ∓ sin A sin B
tantan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)
Double anglesin 2A = 2 sin A cos A; cos 2A = 2 cos²A − 1 = 1 − 2 sin²A
Triple anglesin 3A = 3 sin A − 4 sin³A; cos 3A = 4 cos³A − 3 cos A
Sum to productsin C + sin D = 2 sin((C + D)/2) cos((C − D)/2)
TriangleA + B + C = 180°: tan A + tan B + tan C = tan A tan B tan C

🎯 Most important points for the exam

  • In cos formulas the sign flips, in sin formulas it stays.
  • Name changes at 90° and 270°, stays at 180° and 360°; sign from the quadrant.
  • Double the angle of 3-4-5 and you get 7-24-25.
  • Half-sum and half-difference: sin + sin = 2 sin cos; cos − cos has a minus sign.
  • Triangle identities hold only when A + B + C = 180°.
  • If tan A and tan B are given, look at tan(A + B) first.
Study mantra

Learn the formulas, check the sign first, and verify the answer with numbers once.