Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
'sin A' is one symbol: the sine of the angle A. It is not 'sin' times A.
If you take a bigger right triangle with the same angle A, the triangles are similar, so each ratio stays the same. The ratios belong to the angle, not to a particular triangle.
If one ratio is known, draw a right triangle, find the third side by Pythagoras' theorem, and read off the rest. Example: tan A = 43 gives sides 4k, 3k, hypotenuse 5k, so sin A = 45, cos A = 35.
| ∠A | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin A | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos A | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan A | 0 | 1/√3 | 1 | √3 | not defined |
| cosec A | not defined | 2 | √2 | 2/√3 | 1 |
| sec A | 1 | 2/√3 | √2 | 2 | not defined |
| cot A | not defined | √3 | 1 | 1/√3 | 0 |
sin 0°, 30°, 45°, 60°, 90° are √0/4, √1/4, √2/4, √3/4, √4/4, i.e. 0, ½, 1/√2, √3/2, 1. cos runs the same values backwards.
If tan(A + B) = √3 and tan(A − B) = 1/√3 with A > B, then A + B = 60° and A − B = 30°, so A = 45° and B = 15°.
The first comes from Pythagoras: (opposite)² + (adjacent)² = (hypotenuse)², divided by (hypotenuse)². Dividing it by cos² A or sin² A gives the other two.
| Ratio | in terms of sin A | in terms of cos A |
|---|---|---|
| cos A | √(1 − sin² A) | cos A |
| tan A | sin A / √(1 − sin² A) | √(1 − cos² A) / cos A |
| sec A | 1 / √(1 − sin² A) | 1 / cos A |
Prove (sec A − 1)/(sec A + 1) = (1 − cos A)/(1 + cos A).
| Relation | Formula |
|---|---|
| cosec A | 1/sin A |
| sec A | 1/cos A |
| cot A | 1/tan A |
| tan A | sin A / cos A |
| cot A | cos A / sin A |
Try each problem first, then open the solution to check your working.
In ΔPQR, right-angled at Q, PQ = 8 cm and QR = 6 cm. Find all trigonometric ratios of ∠P.
Given: PQ = 8, QR = 6
Use: PR by Pythagoras
PR = √(64 + 36) = 10
sin P = QR/PR = 3/5, cos P = PQ/PR = 4/5, tan P = 3/4
cosec P = 5/3, sec P = 5/4, cot P = 4/3
Answer: sin 3/5, cos 4/5, tan 3/4, cosec 5/3, sec 5/4, cot 4/3
If sin A = 5/13, find cos A and tan A.
Given: opposite 5k, hypotenuse 13k
adjacent = √(169 − 25)k = 12k
cos A = 12/13, tan A = 5/12
Answer: cos A = 12/13, tan A = 5/12
If tan A = 1√3, find sin A and cos A, and the angle A.
Given: opposite 1, adjacent √3
hypotenuse = √(1 + 3) = 2
sin A = 1/2, cos A = √3/2, so A = 30°
Answer: sin A = 1/2, cos A = √3/2; A = 30°
If cot θ = 4/3, evaluate (sin θ + cos θ)/(sin θ − cos θ).
Use: Divide numerator and denominator by sin θ
(1 + cot θ)/(1 − cot θ) = (7/3)/(−1/3) = −7
Answer: −7
If 5 tan θ = 4, find (5 sin θ − 3 cos θ)/(5 sin θ + 2 cos θ).
Given: tan θ = 4/5
Use: Divide by cos θ
(5 tan θ − 3)/(5 tan θ + 2) = (4 − 3)/(4 + 2)
Answer: 1/6
In ΔABC right-angled at C, AB = 29 cm and BC = 21 cm. Find cos² B − sin² B.
AC = √(841 − 441) = 20
cos B = 21/29, sin B = 20/29
cos² B − sin² B = (441 − 400)/841 = 41/841
Answer: 41/841
Evaluate sin 30° cos 60° + cos 30° sin 60°.
(1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4
Answer: 1
Evaluate (tan² 60° + 4 cos² 45° + 3 sec² 30° + 5 cos² 90°) / (cosec 30° + sec 60° − cot² 30°).
Numerator: 3 + 4(1/2) + 3(4/3) + 0 = 3 + 2 + 4 = 9
Denominator: 2 + 2 − 3 = 1
Answer: 9
If 2 sin 2θ = √3, find θ (0° < 2θ ≤ 90°).
sin 2θ = √3/2 = sin 60°, so 2θ = 60°
Answer: θ = 30°
If sin(A − B) = 1/2 and cos(A + B) = 1/2 with 0° < A + B ≤ 90° and A > B, find A and B.
A − B = 30° and A + B = 60°
Adding: 2A = 90°
Answer: A = 45°, B = 15°
A ladder leaning against a wall makes an angle of 60° with the ground. Its foot is 2.5 m from the wall. Find the length of the ladder.
Given: adjacent = 2.5 m, angle 60°
Use: cos 60° = adjacent / hypotenuse
1/2 = 2.5 / l
Answer: 5 m
Is sin 2A = 2 sin A true for A = 30°?
sin 60° = √3/2 ≈ 0.866; 2 sin 30° = 1
Answer: No
Simplify (1 − cos² A) cosec² A.
Use: 1 − cos² A = sin² A
sin² A × 1/sin² A
Answer: 1
Simplify (1 + tan² θ)(1 − sin θ)(1 + sin θ).
sec² θ × (1 − sin² θ) = sec² θ × cos² θ
Answer: 1
Prove that (1 − sin θ)/cos θ = cos θ/(1 + sin θ).
Use: Multiply by (1 + sin θ)/(1 + sin θ)
LHS = (1 − sin² θ)/(cos θ (1 + sin θ)) = cos² θ/(cos θ (1 + sin θ)) = cos θ/(1 + sin θ)
Answer: Proved
Prove that sec⁴ A − sec² A = tan⁴ A + tan² A.
Use: sec² A = 1 + tan² A
LHS = sec² A (sec² A − 1) = (1 + tan² A) tan² A = tan² A + tan⁴ A
Answer: Proved
Prove that cos A/(1 − tan A) + sin A/(1 − cot A) = sin A + cos A.
Use: Write in sin and cos
cos A/((cos A − sin A)/cos A) + sin A/((sin A − cos A)/sin A)
= (cos² A − sin² A)/(cos A − sin A) = cos A + sin A
Answer: Proved
If sin θ + cos θ = √2 cos θ, show that cos θ − sin θ = √2 sin θ.
sin θ = (√2 − 1) cos θ, so cos θ = sin θ/(√2 − 1) = (√2 + 1) sin θ
cos θ − sin θ = √2 sin θ
Answer: Proved
If sec θ + tan θ = 3, find sec θ − tan θ and hence sec θ.
Use: sec² θ − tan² θ = 1
(sec θ + tan θ)(sec θ − tan θ) = 1 ⇒ sec θ − tan θ = 1/3
Adding: 2 sec θ = 10/3
Answer: sec θ − tan θ = 1/3; sec θ = 5/3
If x = a sec θ and y = b tan θ, show that x²/a² − y²/b² = 1.
x²/a² − y²/b² = sec² θ − tan² θ
Answer: = 1
Pick a level and answer 10 questions with instant feedback. Each level has a bank of 20 questions, and New 10 questions draws a fresh set.
Read the Assertion (A) and the Reason (R), then pick one of the four standard options. You get 5 questions per set from a bank of 20 at each level, and New 5 questions draws a fresh set.
Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
AC² = 24² + 7² = 576 + 49 = 625, so AC = 25 cm.
(i) sin A = BC/AC = 7/25, cos A = AB/AC = 24/25.
(ii) sin C = AB/AC = 24/25, cos C = BC/AC = 7/25.
Answer: (i) 7/25, 24/25 (ii) 24/25, 7/25
QR = √(13² − 12²) = √25 = 5 cm.
tan P = QR/PQ = 5/12 and cot R = QR/PQ = 5/12.
tan P − cot R = 0.
Answer: 0
Take opposite = 3k and hypotenuse = 4k. Adjacent = √(16k² − 9k²) = √7 k.
cos A = √74, tan A = 3√7.
Answer: cos A = √7/4, tan A = 3/√7
cot A = 8/15: adjacent = 8k, opposite = 15k, hypotenuse = √(64 + 225)k = 17k.
sin A = 15/17, sec A = 17/8.
Answer: sin A = 15/17, sec A = 17/8
cos θ = 12/13: adjacent 12k, hypotenuse 13k, opposite = √(169 − 144)k = 5k.
sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cot θ = 12/5, cosec θ = 13/5.
Answer: sin 5/13, cos 12/13, tan 5/12, cot 12/5, cosec 13/5
Take a right triangle ABC with the right angle at C. Then cos A = AC/AB and cos B = BC/AB.
cos A = cos B ⇒ AC/AB = BC/AB ⇒ AC = BC.
Angles opposite equal sides are equal, so ∠B = ∠A.
(Equivalently, the ratios depend only on the angle; equal cosines of acute angles give equal angles.)
Answer: Proved
(i) 1 − sin² θ1 − cos² θ = cos² θsin² θ = cot² θ = 4964.
(ii) cot² θ = (7/8)² = 49/64.
Answer: (i) 49/64 (ii) 49/64
cot A = 4/3, so tan A = 3/4. Sides: adjacent 4k, opposite 3k, hypotenuse 5k.
LHS = 1 − 9/161 + 9/16 = 7/1625/16 = 725.
RHS = (4/5)² − (3/5)² = 16 − 925 = 725.
Answer: Yes, both sides equal 7/25
tan A = 1/√3: BC = k, AB = √3 k, AC = 2k. So sin A = 1/2, cos A = √3/2, sin C = √3/2, cos C = 1/2.
(i) (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.
(ii) (√3/2)(1/2) − (1/2)(√3/2) = 0.
Answer: (i) 1 (ii) 0
Let QR = x; then PR = 25 − x. By Pythagoras, (25 − x)² = 25 + x² ⇒ 625 − 50x = 25 ⇒ x = 12.
So QR = 12 cm, PR = 13 cm.
sin P = QR/PR = 12/13, cos P = PQ/PR = 5/13, tan P = QR/PQ = 12/5.
Answer: sin P = 12/13, cos P = 5/13, tan P = 12/5
(i) False. tan 60° = √3 > 1.
(ii) True. sec A = hypotenuse/adjacent ≥ 1, and 12/5 > 1 (e.g. a triangle with sides 5, √119, 12).
(iii) False. cos A is cosine of A; cosecant is written cosec A.
(iv) False. cot A is a single symbol for the cotangent of A.
(v) False. sin θ = opposite/hypotenuse < 1 for an acute angle, so it cannot be 4/3.
Answer: (i) F (ii) T (iii) F (iv) F (v) F
(i) (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1.
(ii) 2(1)² + (√3/2)² − (√3/2)² = 2.
(iii) 1/√22/√3 + 2 = 1/√2(2 + 2√3)/√3 = √32√2(1 + √3). Rationalising: √3(√3 − 1)2√2 × 2 = 3 − √34√2 = 3√2 − √68.
(iv) 1/2 + 1 − 2/√32/√3 + 1/2 + 1 = 3√3 − 43√3 + 4 (multiplying by 2√3) = (3√3 − 4)²27 − 16 = 43 − 24√311.
(v) Numerator: 5(1/4) + 4(4/3) − 1 = 5/4 + 16/3 − 1 = 6712. Denominator: 1/4 + 3/4 = 1. Value 6712.
Answer: (i) 1 (ii) 2 (iii) (3√2 − √6)/8 (iv) (43 − 24√3)/11 (v) 67/12
(i) (2/√3)/(1 + 1/3) = (2/√3)(3/4) = √3/2 = sin 60°: (A).
(ii) (1 − 1)/(1 + 1) = 0: (D).
(iii) At A = 0°, both sides are 0. At 30°: sin 60° = √3/2 ≠ 1. At 45°: 1 ≠ √2. At 60°: 0 ≠ √3. So (A).
(iv) (2/√3)/(1 − 1/3) = (2/√3)(3/2) = √3 = tan 60°: (C).
Answer: (i) A (ii) D (iii) A (iv) C
tan (A + B) = √3 ⇒ A + B = 60°. tan (A − B) = 1/√3 ⇒ A − B = 30°.
Adding: 2A = 90°, A = 45°; then B = 15°.
Answer: A = 45°, B = 15°
(i) False. For A = B = 30°: sin 60° = √3/2 but sin 30° + sin 30° = 1.
(ii) True (for 0° ≤ θ ≤ 90°): 0, 1/2, 1/√2, √3/2, 1.
(iii) False. cos θ decreases: 1, √3/2, 1/√2, 1/2, 0.
(iv) False. They are equal only at θ = 45°; e.g. sin 30° ≠ cos 30°.
(v) True. cot 0° = cos 0°/sin 0° = 1/0, which is not defined.
Answer: (i) F (ii) T (iii) F (iv) F (v) T
cosec² A = 1 + cot² A, so sin A = 1√1 + cot² A.
tan A = 1cot A.
sec² A = 1 + tan² A = 1 + 1cot² A = cot² A + 1cot² A, so sec A = √1 + cot² Acot A.
Answer: sin A = 1/√(1 + cot² A); sec A = √(1 + cot² A)/cot A; tan A = 1/cot A
cos A = 1sec A
tan A = √sec² A − 1
sin A = tan A · cos A = √sec² A − 1sec A
cot A = 1√sec² A − 1
cosec A = sec A√sec² A − 1
Answer: as listed
(i) 9(sec² A − tan² A) = 9 × 1 = 9: (B).
(ii) = cos θ + sin θ + 1cos θ × sin θ + cos θ − 1sin θ = (sin θ + cos θ)² − 1sin θ cos θ = 2 sin θ cos θsin θ cos θ = 2: (C).
(iii) = 1 + sin Acos A × (1 − sin A) = 1 − sin² Acos A = cos² Acos A = cos A: (D).
(iv) = sec² Acosec² A = sin² Acos² A = tan² A: (D).
Answer: (i) B (ii) C (iii) D (iv) D
(i) LHS = (1sin θ − cos θsin θ)² = (1 − cos θ)²sin² θ = (1 − cos θ)²(1 − cos θ)(1 + cos θ) = 1 − cos θ1 + cos θ = RHS.
(ii) LHS = cos² A + (1 + sin A)²(1 + sin A) cos A = cos² A + sin² A + 1 + 2 sin A(1 + sin A) cos A = 2(1 + sin A)(1 + sin A) cos A = 2cos A = 2 sec A.
(iii) With s = sin θ, c = cos θ: LHS = s/c(s − c)/s + c/s(c − s)/c = s²c(s − c) − c²s(s − c) = s³ − c³sc(s − c) = s² + sc + c²sc = 1 + scsc = 1 + 1sc = 1 + sec θ cosec θ.
(iv) LHS = 1sec A + 1 = cos A + 1. RHS = 1 − cos² A1 − cos A = (1 − cos A)(1 + cos A)1 − cos A = 1 + cos A. So LHS = RHS.
(v) Divide numerator and denominator by sin A: LHS = cot A − 1 + cosec Acot A + 1 − cosec A. Replace 1 in the numerator by cosec² A − cot² A: numerator = (cot A + cosec A) − (cosec A − cot A)(cosec A + cot A) = (cot A + cosec A)(1 − cosec A + cot A). This last bracket equals the denominator, so LHS = cosec A + cot A.
(vi) LHS = √(1 + sin A)²(1 − sin A)(1 + sin A) = √(1 + sin A)²cos² A = 1 + sin Acos A = sec A + tan A.
(vii) LHS = sin θ (1 − 2 sin² θ)cos θ (2 cos² θ − 1). Since 1 − 2 sin² θ = 1 − 2(1 − cos² θ) = 2 cos² θ − 1, LHS = sin θcos θ = tan θ.
(viii) LHS = sin² A + cosec² A + 2 + cos² A + sec² A + 2 = (sin² A + cos² A) + 4 + (1 + cot² A) + (1 + tan² A) = 7 + tan² A + cot² A.
(ix) LHS = 1 − sin² Asin A × 1 − cos² Acos A = cos² A sin² Asin A cos A = sin A cos A. RHS = 1(sin² A + cos² A)/(sin A cos A) = sin A cos A.
(x) 1 + tan² A1 + cot² A = sec² Acosec² A = sin² Acos² A = tan² A. Also 1 − tan A1 − cot A = 1 − tan A(tan A − 1)/tan A = −tan A, so its square is tan² A.
Answer: All ten identities proved