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You Won't Believe How SIMPLE Trigonometry Can Be in CBSE Class 10
Prefer to watch before you read? This video walks through the chapter's core ideas — trigonometric ratios, the standard angles, and how the six ratios relate to each other — before you work through the solved questions below.
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Key Concepts & Formulae at a Glance
- Six ratios:sin A = opphyp, cos A = adjhyp, tan A = oppadj, with cosec A, sec A, cot A as their reciprocals.
- tan A and cot A:tan A = sin Acos A, and cot A = cos Asin A = 1tan A.
- Ratios are angle-dependent, not size-dependent:Similar right triangles (same acute angle) give identical trigonometric ratios, however large or small the triangle.
- Table of standard angles:sin/cos/tan of 0°, 30°, 45°, 60°, 90° are fixed, exact values — the foundation for every numeric problem in this chapter.
- Three Pythagorean identities:sin²A + cos²A = 1 (always); 1 + tan²A = sec²A (for A ≠ 90°); 1 + cot²A = cosec²A (for A ≠ 0°).
- Range of values:sin A and cos A never exceed 1; sec A and cosec A are never less than 1 (in magnitude).
Exercise 8.1
11 Questions
1In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
ΔABC, right-angled at B — AC found from the Pythagoras theorem.
AC² = AB²+BC² = 24²+7² = 576+49 = 625 (Pythagoras theorem)
AC = 25 cm
For ∠A: opposite = BC = 7, adjacent = AB = 24, hypotenuse = AC = 25
sin A = BCAC = 725, cos A = ABAC = 2425
For ∠C: opposite = AB = 24, adjacent = BC = 7, hypotenuse = AC = 25
sin C = ABAC = 2425, cos C = BCAC = 725
AC = 25 cm
For ∠A: opposite = BC = 7, adjacent = AB = 24, hypotenuse = AC = 25
sin A = BCAC = 725, cos A = ABAC = 2425
For ∠C: opposite = AB = 24, adjacent = BC = 7, hypotenuse = AC = 25
sin C = ABAC = 2425, cos C = BCAC = 725
sin A = 725, cos A = 2425; sin C = 2425, cos C = 725
2In Fig. 8.13, find tan P − cot R.
ΔPQR, right-angled at Q — QR found from the Pythagoras theorem.
ΔPQR is right-angled at Q, with PQ = 12 cm, PR = 13 cm.
QR² = PR²−PQ² = 169−144 = 25 (Pythagoras theorem)
QR = 5 cm
tan P = oppositeadjacent (for ∠P) = QRPQ = 512
cot R = adjacentopposite (for ∠R) = QRPQ = 512 (for ∠R, opposite is PQ, adjacent is QR)
tan P − cot R = 512 − 512
QR² = PR²−PQ² = 169−144 = 25 (Pythagoras theorem)
QR = 5 cm
tan P = oppositeadjacent (for ∠P) = QRPQ = 512
cot R = adjacentopposite (for ∠R) = QRPQ = 512 (for ∠R, opposite is PQ, adjacent is QR)
tan P − cot R = 512 − 512
tan P − cot R = 0
3If sin A = 34, calculate cos A and tan A.
sin A = oppositehypotenuse = 34, so let opposite (BC) = 3k, hypotenuse (AC) = 4k.
AB² = AC²−BC² = (4k)²−(3k)² = 16k²−9k² = 7k² (Pythagoras theorem)
AB = k√7
cos A = ABAC = k√74k = √74
tan A = BCAB = 3kk√7 = 3√7 = 3√77
AB² = AC²−BC² = (4k)²−(3k)² = 16k²−9k² = 7k² (Pythagoras theorem)
AB = k√7
cos A = ABAC = k√74k = √74
tan A = BCAB = 3kk√7 = 3√7 = 3√77
cos A = √74, tan A = 3√77
4Given 15 cot A = 8, find sin A and sec A.
cot A = 815, so adjacent (AB) = 8k, opposite (BC) = 15k.
AC² = AB²+BC² = (8k)²+(15k)² = 64k²+225k² = 289k² (Pythagoras theorem)
AC = 17k
sin A = BCAC = 15k17k = 1517
sec A = ACAB = 17k8k = 178
AC² = AB²+BC² = (8k)²+(15k)² = 64k²+225k² = 289k² (Pythagoras theorem)
AC = 17k
sin A = BCAC = 15k17k = 1517
sec A = ACAB = 17k8k = 178
sin A = 1517, sec A = 178
5Given sec θ = 1312, calculate all other trigonometric ratios.
sec θ = hypotenuseadjacent = 1312, so hypotenuse (AC) = 13k, adjacent (AB) = 12k.
BC² = AC²−AB² = (13k)²−(12k)² = 169k²−144k² = 25k² (Pythagoras theorem)
BC = 5k
sin θ = BCAC = 513, cos θ = ABAC = 1213, tan θ = BCAB = 512
cosec θ = 1sinθ = 135, cot θ = 1tanθ = 125
BC² = AC²−AB² = (13k)²−(12k)² = 169k²−144k² = 25k² (Pythagoras theorem)
BC = 5k
sin θ = BCAC = 513, cos θ = ABAC = 1213, tan θ = BCAB = 512
cosec θ = 1sinθ = 135, cot θ = 1tanθ = 125
sinθ=513, cosθ=1213, tanθ=512, cosecθ=135, secθ=1312, cotθ=125
6If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Consider a triangle ABC in which CD ⊥ AB, with D on AB (see the construction below).
In right ΔCAD: cos A = ADAC. In right ΔCBD: cos B = BDBC.
Given cos A = cos B: ADAC = BDBC ⇒ ADBD = ACBC = k (say)
So AD = k·BD and AC = k·BC.
In ΔCAD, CD² = AC²−AD² = k²BC²−k²BD² = k²(BC²−BD²) (Pythagoras theorem)
But in ΔCBD, BC²−BD² = CD² (Pythagoras theorem)
So CD² = k²CD² ⇒ k²=1 ⇒ k=1 (k is a positive ratio)
With k=1, AD = BD. Now in ΔCAD and ΔCBD: AD=BD, CD=CD (common), ∠ADC=∠BDC=90°
So ΔCAD ≅ ΔCBD (SAS congruency)
Therefore ∠A = ∠B (CPCT)
In right ΔCAD: cos A = ADAC. In right ΔCBD: cos B = BDBC.
Given cos A = cos B: ADAC = BDBC ⇒ ADBD = ACBC = k (say)
So AD = k·BD and AC = k·BC.
In ΔCAD, CD² = AC²−AD² = k²BC²−k²BD² = k²(BC²−BD²) (Pythagoras theorem)
But in ΔCBD, BC²−BD² = CD² (Pythagoras theorem)
So CD² = k²CD² ⇒ k²=1 ⇒ k=1 (k is a positive ratio)
With k=1, AD = BD. Now in ΔCAD and ΔCBD: AD=BD, CD=CD (common), ∠ADC=∠BDC=90°
So ΔCAD ≅ ΔCBD (SAS congruency)
Therefore ∠A = ∠B (CPCT)
∠A = ∠B, as required.
7If cot θ = 78, evaluate:
(i) (1+sinθ)(1−sinθ) / [(1+cosθ)(1−cosθ)]
(ii) cot²θ
cot θ = 78, so adjacent (AB) = 7k, opposite (BC) = 8k.
AC² = 7²k²+8²k² = 113k² (Pythagoras theorem) ⇒ AC = k√113
sinθ = 8√113, cosθ = 7√113
(i) (1+sinθ)(1−sinθ) = 1−sin²θ = cos²θ (identity), and similarly (1+cosθ)(1−cosθ) = sin²θ. So the expression simplifies to cos²θsin²θ = cot²θ = (78)² = 4964.
(ii) cot²θ = (78)² = 4964
AC² = 7²k²+8²k² = 113k² (Pythagoras theorem) ⇒ AC = k√113
sinθ = 8√113, cosθ = 7√113
(i) (1+sinθ)(1−sinθ) = 1−sin²θ = cos²θ (identity), and similarly (1+cosθ)(1−cosθ) = sin²θ. So the expression simplifies to cos²θsin²θ = cot²θ = (78)² = 4964.
(ii) cot²θ = (78)² = 4964
(i) 4964 (ii) 4964 — both expressions equal cot²θ, so they come out identical.
8If 3 cot A = 4, check whether 1−tan²A1+tan²A = cos²A − sin²A or not.
cot A = 43, so tan A = 34. Let opposite (BC) = 3k, adjacent (AB) = 4k.
AC² = (4k)²+(3k)² = 25k² ⇒ AC = 5k (Pythagoras theorem)
sin A = 35, cos A = 45
LHS: 1−tan²A1+tan²A = (1−916)/(1+916) = (716)/(2516) = 725
RHS: cos²A−sin²A = 1625−925 = 725
AC² = (4k)²+(3k)² = 25k² ⇒ AC = 5k (Pythagoras theorem)
sin A = 35, cos A = 45
LHS: 1−tan²A1+tan²A = (1−916)/(1+916) = (716)/(2516) = 725
RHS: cos²A−sin²A = 1625−925 = 725
Yes, LHS = RHS = 725 — the identity holds for this value of A.
9In triangle ABC, right-angled at B, if tan A = 1√3, find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C − sin A sin C
tan A = 1√3, so opposite (BC) = k, adjacent (AB) = √3k.
AC² = 3k²+k² = 4k² ⇒ AC = 2k (Pythagoras theorem)
sin A = BCAC = 12, cos A = ABAC = √32
Since ∠A+∠C = 90° (angle sum, right angle at B), ∠C = 90°−A, so sin C = cos A = √32 and cos C = sin A = 12.
(i) sin A cos C + cos A sin C = (12)(12) + (√32)(√32) = 14+34
(ii) cos A cos C − sin A sin C = (√32)(12) − (12)(√32) = √34 − √34
AC² = 3k²+k² = 4k² ⇒ AC = 2k (Pythagoras theorem)
sin A = BCAC = 12, cos A = ABAC = √32
Since ∠A+∠C = 90° (angle sum, right angle at B), ∠C = 90°−A, so sin C = cos A = √32 and cos C = sin A = 12.
(i) sin A cos C + cos A sin C = (12)(12) + (√32)(√32) = 14+34
(ii) cos A cos C − sin A sin C = (√32)(12) − (12)(√32) = √34 − √34
(i) 1 (ii) 0
10In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Let QR = x, so PR = 25−x.
PR² = PQ²+QR² (Pythagoras theorem)
(25−x)² = 5²+x²
625−50x+x² = 25+x²
600 = 50x
x = 12, so QR = 12 cm and PR = 25−12 = 13 cm
For ∠P: opposite = QR = 12, adjacent = PQ = 5, hypotenuse = PR = 13
sin P = QRPR = 1213, cos P = PQPR = 513, tan P = QRPQ = 125
PR² = PQ²+QR² (Pythagoras theorem)
(25−x)² = 5²+x²
625−50x+x² = 25+x²
600 = 50x
x = 12, so QR = 12 cm and PR = 25−12 = 13 cm
For ∠P: opposite = QR = 12, adjacent = PQ = 5, hypotenuse = PR = 13
sin P = QRPR = 1213, cos P = PQPR = 513, tan P = QRPQ = 125
sin P = 1213, cos P = 513, tan P = 125
11State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 125 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 43 for some angle θ.
(i) False
For example, tan 60° = √3 ≈ 1.73, which is greater than 1. tan A depends on the specific angle and can be less than, equal to, or greater than 1.
(ii) True
sec A ≥ 1 always (for acute A), and 125 = 2.4 ≥ 1, so it's a valid value. A right triangle with hypotenuse 12 and adjacent side 5 (and opposite √119) realizes this exactly.
(iii) False
"cos A" is the abbreviation for the cosine of A. The cosecant of A is abbreviated "cosec A".
(iv) False
"cot A" is a single symbol meaning "the cotangent of the angle A" — it is not the product of a separate quantity "cot" and the number A. The same convention applies to sin A, cos A, tan A, etc.
(v) False
sin θ can never exceed 1, since the opposite side of a right triangle can never be longer than the hypotenuse. As 43 > 1, no angle θ gives sin θ = 43.
Exercise 8.2
4 Questions
1Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan²45° + cos²30° − sin²60°
(iii) cos45° / (sec30° + cosec30°)
(iv) (sin30°+tan45°−cosec60°) / (sec30°+cos60°+cot45°)
(v) (5cos²60°+4sec²30°−tan²45°) / (sin²30°+cos²30°)
(i)
sin60°cos30°+sin30°cos60° = (√32)(√32)+(12)(12) = 34+14
1
(ii)
2tan²45°+cos²30°−sin²60° = 2(1)²+(√32)²−(√32)² = 2+34−34
2
(iii)
cos45° = 1√2; sec30°=2√3, cosec30°=2, so sec30°+cosec30° = 2√3+2 = 21+√3√3
Expression = (1√2) ÷ [21+√3√3] = √3/[2√2(1+√3)]
Rationalising (multiply num & denom by (√3−1)): = √3(√3−1)/[2√2×2] = 3−√34√2 = 3√2−√68
Expression = (1√2) ÷ [21+√3√3] = √3/[2√2(1+√3)]
Rationalising (multiply num & denom by (√3−1)): = √3(√3−1)/[2√2×2] = 3−√34√2 = 3√2−√68
3√2 − √68
(iv)
Numerator = sin30°+tan45°−cosec60° = 12+1−2√3 = 32−2√3
Denominator = sec30°+cos60°+cot45° = 2√3+12+1 = 2√3+32
Multiplying through and simplifying gives 43−24√311
Denominator = sec30°+cos60°+cot45° = 2√3+12+1 = 2√3+32
Multiplying through and simplifying gives 43−24√311
43 − 24√311 ≈ 0.130
(v)
Numerator = 5(12)²+4(2√3)²−1² = 54+163−1 = 1512+6412−1212 = 6712
Denominator = sin²30°+cos²30° = 1 (identity, true for any angle)
Denominator = sin²30°+cos²30° = 1 (identity, true for any angle)
6712
2Choose the correct option and justify your choice:
(i) 2tan30°1+tan²30° = (A) sin60° (B) cos60° (C) tan60° (D) sin30°
(ii) 1−tan²45°1+tan²45° = (A) tan90° (B) 1 (C) sin45° (D) 0
(iii) sin2A = 2sinA is true when A = (A) 0° (B) 30° (C) 45° (D) 60°
(iv) 2tan30°1−tan²30° = (A) cos60° (B) sin60° (C) tan60° (D) sin30°
(i)
2tan30°1+tan²30° = 2(1√3)/(1+13) = (2√3)/(43) = (2√3)×(34) = 64√3 = √32 = sin60°
(A) sin 60°
(ii)
1−tan²45°1+tan²45° = 1−11+1 = 02
(D) 0
(iii)
At A=0°: sin2A=sin0°=0, and 2sinA=2sin0°=0 — both sides equal 0. (For other listed options, e.g. A=30°: sin60°=√32 but 2sin30°=1, not equal.)
(A) 0°
(iv)
2tan30°1−tan²30° = 2(1√3)/(1−13) = (2√3)/(23) = (2√3)×(32) = 3√3 = √3 = tan60°
(C) tan 60°
3If tan(A+B) = √3 and tan(A−B) = 1√3; 0° < A+B ≤ 90°; A > B, find A and B.
tan(A+B) = √3 = tan60° ⇒ A+B = 60° ...(1)
tan(A−B) = 1√3 = tan30° ⇒ A−B = 30° ...(2)
Adding (1) and (2): 2A = 90° ⇒ A = 45°
Substituting into (1): B = 60°−45° = 15°
tan(A−B) = 1√3 = tan30° ⇒ A−B = 30° ...(2)
Adding (1) and (2): 2A = 90° ⇒ A = 45°
Substituting into (1): B = 60°−45° = 15°
A = 45°, B = 15°
4State whether the following are true or false. Justify your answer.
(i) sin(A+B) = sinA + sinB.
(ii) The value of sinθ increases as θ increases.
(iii) The value of cosθ increases as θ increases.
(iv) sinθ = cosθ for all values of θ.
(v) cot A is not defined for A = 0°.
(i) False
Take A=B=30°: sin(A+B)=sin60°=√32≈0.866, but sinA+sinB=sin30°+sin30°=12+12=1. Since 0.866≠1, the equation fails in general.
(ii) True
From the table of standard angles, sinθ rises steadily from 0 (at 0°) to 1 (at 90°) — sine is an increasing function over this range.
(iii) False
cosθ falls from 1 (at 0°) to 0 (at 90°) — cosine is a decreasing function over this range, the opposite of increasing.
(iv) False
sinθ=cosθ only at θ=45° (where both equal 1√2). At any other angle, e.g. θ=30°: sin30°=12 but cos30°=√32 — not equal.
(v) True
cot A = cosAsinA, and sin0°=0. Division by zero is undefined, so cot 0° is not defined.
Exercise 8.3
4 Questions
1Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Starting from cosec²A = 1+cot²A: cosec A = √(1+cot²A), so sin A = 1cosec A = 1√(1+cot²A)
tan A = 1cot A
sec A = √(1+tan²A) = √(1+1cot²A) = √cot²A+1cot A
tan A = 1cot A
sec A = √(1+tan²A) = √(1+1cot²A) = √cot²A+1cot A
sin A = 1√(1+cot²A); tan A = 1cot A; sec A = √1+cot²Acot A
2Write all the other trigonometric ratios of ∠A in terms of sec A.
cos A = 1sec A
From sec²A = 1+tan²A: tan A = √(sec²A−1)
cot A = 1tan A = 1√(sec²A−1)
From sin²A = 1−cos²A = 1−1sec²A = sec²A−1sec²A: sin A = √sec²A−1sec A
cosec A = 1sin A = sec A√(sec²A−1)
From sec²A = 1+tan²A: tan A = √(sec²A−1)
cot A = 1tan A = 1√(sec²A−1)
From sin²A = 1−cos²A = 1−1sec²A = sec²A−1sec²A: sin A = √sec²A−1sec A
cosec A = 1sin A = sec A√(sec²A−1)
cos A = 1secA; sinA = √sec²A−1secA; tanA = √(sec²A−1); cotA = 1√(sec²A−1); cosecA = secA√(sec²A−1)
3Choose the correct option. Justify your choice:
(i) 9sec²A − 9tan²A = (A) 1 (B) 9 (C) 8 (D) 0
(ii) (1+tanθ+secθ)(1+cotθ−cosecθ) = (A) 0 (B) 1 (C) 2 (D) −1
(iii) (secA+tanA)(1−sinA) = (A) secA (B) sinA (C) cosecA (D) cosA
(iv) 1+tan²A1+cot²A = (A) sec²A (B) −1 (C) cot²A (D) tan²A
(i)
9sec²A−9tan²A = 9(sec²A−tan²A) = 9(1) (identity: sec²A−tan²A=1)
(B) 9
(ii)
Writing everything in sinθ, cosθ: (1+tanθ+secθ) = cosθ+sinθ+1cosθ, and (1+cotθ−cosecθ) = sinθ+cosθ−1sinθ.
Product = (cosθ+sinθ+1)sinθ+cosθ−1sinθcosθ = [(sinθ+cosθ)²−1]/(sinθcosθ)
(sinθ+cosθ)² = sin²θ+cos²θ+2sinθcosθ = 1+2sinθcosθ, so numerator = 1+2sinθcosθ−1 = 2sinθcosθ
Product = 2sinθcosθsinθcosθ = 2
Product = (cosθ+sinθ+1)sinθ+cosθ−1sinθcosθ = [(sinθ+cosθ)²−1]/(sinθcosθ)
(sinθ+cosθ)² = sin²θ+cos²θ+2sinθcosθ = 1+2sinθcosθ, so numerator = 1+2sinθcosθ−1 = 2sinθcosθ
Product = 2sinθcosθsinθcosθ = 2
(C) 2
(iii)
(secA+tanA)(1−sinA) = (1cosA+sinAcosA)(1−sinA) = [1+sinAcosA](1−sinA) = 1−sin²AcosA = cos²AcosA
(D) cos A
(iv)
1+tan²A1+cot²A = sec²Acosec²A = (1cos²A)/(1sin²A) = sin²Acos²A
(D) tan²A
4Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
(i) (cosecθ−cotθ)² = 1−cosθ1+cosθ
(ii) cosA1+sinA + 1+sinAcosA = 2secA
(iii) tanθ1−cotθ + cotθ1−tanθ = 1+secθcosecθ
(iv) 1+secAsecA = sin²A1−cosA
(v) cosA−sinA+1cosA+sinA−1 = cosecA+cotA
(vi) √[1+sinA1−sinA] = secA+tanA
(vii) sinθ−2sin³θ2cos³θ−cosθ = tanθ
(viii) (sinA+cosecA)²+(cosA+secA)² = 7+tan²A+cot²A
(ix) (cosecA−sinA)(secA−cosA) = 1tanA+cotA
(x) 1+tan²A1+cot²A = [1−tanA1−cotA]² = tan²A
(i)
LHS = (cosecθ−cotθ)² = (1sinθ − cosθsinθ)² = [1−cosθsinθ]² = (1−cosθ)²sin²θ
= (1−cosθ)²1−cos²θ = (1−cosθ)²/[(1−cosθ)(1+cosθ)] = 1−cosθ1+cosθ = RHS
= (1−cosθ)²1−cos²θ = (1−cosθ)²/[(1−cosθ)(1+cosθ)] = 1−cosθ1+cosθ = RHS
Proved.
(ii)
LHS = [cos²A + (1+sinA)²] / [cosA(1+sinA)]
Numerator = cos²A+1+2sinA+sin²A = (sin²A+cos²A)+1+2sinA = 1+1+2sinA = 2+2sinA = 2(1+sinA)
LHS = 2(1+sinA)/[cosA(1+sinA)] = 2cosA = 2secA = RHS
Numerator = cos²A+1+2sinA+sin²A = (sin²A+cos²A)+1+2sinA = 1+1+2sinA = 2+2sinA = 2(1+sinA)
LHS = 2(1+sinA)/[cosA(1+sinA)] = 2cosA = 2secA = RHS
Proved.
(iii)
Writing tanθ=sinθcosθ, cotθ=cosθsinθ:
tanθ1−cotθ = (sinθcosθ)/(1−cosθsinθ) = (sinθcosθ)·[sinθsinθ−cosθ] = sin²θ/[cosθ(sinθ−cosθ)]
cotθ1−tanθ = (cosθsinθ)/(1−sinθcosθ) = (cosθsinθ)·[cosθcosθ−sinθ] = −cos²θ/[sinθ(sinθ−cosθ)]
Sum = [sin²θ·sinθ − cos²θ·cosθ] / [sinθcosθ(sinθ−cosθ)] = (sin³θ−cos³θ)/[sinθcosθ(sinθ−cosθ)]
Using a³−b³=(a−b)(a²+ab+b²): sin³θ−cos³θ = (sinθ−cosθ)(sin²θ+sinθcosθ+cos²θ) = (sinθ−cosθ)(1+sinθcosθ)
Sum = 1+sinθcosθsinθcosθ = 1sinθcosθ + 1 = secθcosecθ+1 = RHS
tanθ1−cotθ = (sinθcosθ)/(1−cosθsinθ) = (sinθcosθ)·[sinθsinθ−cosθ] = sin²θ/[cosθ(sinθ−cosθ)]
cotθ1−tanθ = (cosθsinθ)/(1−sinθcosθ) = (cosθsinθ)·[cosθcosθ−sinθ] = −cos²θ/[sinθ(sinθ−cosθ)]
Sum = [sin²θ·sinθ − cos²θ·cosθ] / [sinθcosθ(sinθ−cosθ)] = (sin³θ−cos³θ)/[sinθcosθ(sinθ−cosθ)]
Using a³−b³=(a−b)(a²+ab+b²): sin³θ−cos³θ = (sinθ−cosθ)(sin²θ+sinθcosθ+cos²θ) = (sinθ−cosθ)(1+sinθcosθ)
Sum = 1+sinθcosθsinθcosθ = 1sinθcosθ + 1 = secθcosecθ+1 = RHS
Proved.
(iv)
LHS = 1+secAsecA = 1secA + 1 = cosA+1
RHS = sin²A1−cosA = 1−cos²A1−cosA = (1−cosA)1+cosA1−cosA = 1+cosA
RHS = sin²A1−cosA = 1−cos²A1−cosA = (1−cosA)1+cosA1−cosA = 1+cosA
LHS = RHS = 1+cosA. Proved.
(v)
Divide numerator and denominator of LHS by sinA:
LHS = cotA−1+cosecAcotA+1−cosecA = [(cotA+cosecA)−1] / [1−(cosecA−cotA)]
Using cosec²A−cot²A=1, i.e. (cosecA−cotA)(cosecA+cotA)=1: cosecA−cotA = 1cosecA+cotA
So the denominator becomes 1−1cosecA+cotA = [(cosecA+cotA)−1] / (cosecA+cotA)
LHS = [(cotA+cosecA)−1] ÷ {[(cosecA+cotA)−1]/(cosecA+cotA)} = cosecA+cotA = RHS
LHS = cotA−1+cosecAcotA+1−cosecA = [(cotA+cosecA)−1] / [1−(cosecA−cotA)]
Using cosec²A−cot²A=1, i.e. (cosecA−cotA)(cosecA+cotA)=1: cosecA−cotA = 1cosecA+cotA
So the denominator becomes 1−1cosecA+cotA = [(cosecA+cotA)−1] / (cosecA+cotA)
LHS = [(cotA+cosecA)−1] ÷ {[(cosecA+cotA)−1]/(cosecA+cotA)} = cosecA+cotA = RHS
Proved.
(vi)
LHS = √[1+sinA1−sinA] = √{[(1+sinA)(1+sinA)]/[(1−sinA)(1+sinA)]} = √[(1+sinA)²1−sin²A]
= √[(1+sinA)²cos²A] = 1+sinAcosA (cosA>0 for acute A)
= 1cosA + sinAcosA = secA+tanA = RHS
= √[(1+sinA)²cos²A] = 1+sinAcosA (cosA>0 for acute A)
= 1cosA + sinAcosA = secA+tanA = RHS
Proved.
(vii)
LHS = [sinθ(1−2sin²θ)] / [cosθ(2cos²θ−1)]
Using sin²θ=1−cos²θ: 1−2sin²θ = 1−2(1−cos²θ) = 2cos²θ−1 (same expression as the denominator's bracket)
LHS = [sinθ(2cos²θ−1)] / [cosθ(2cos²θ−1)] = sinθcosθ = tanθ = RHS
Using sin²θ=1−cos²θ: 1−2sin²θ = 1−2(1−cos²θ) = 2cos²θ−1 (same expression as the denominator's bracket)
LHS = [sinθ(2cos²θ−1)] / [cosθ(2cos²θ−1)] = sinθcosθ = tanθ = RHS
Proved.
(viii)
LHS = (sinA+cosecA)²+(cosA+secA)²
= sin²A+2sinA·cosecA+cosec²A + cos²A+2cosA·secA+sec²A
= (sin²A+cos²A) + 2(sinA·1sinA) + 2(cosA·1cosA) + cosec²A+sec²A
= 1+2+2+cosec²A+sec²A = 5+(1+cot²A)+(1+tan²A) = 7+tan²A+cot²A = RHS
= sin²A+2sinA·cosecA+cosec²A + cos²A+2cosA·secA+sec²A
= (sin²A+cos²A) + 2(sinA·1sinA) + 2(cosA·1cosA) + cosec²A+sec²A
= 1+2+2+cosec²A+sec²A = 5+(1+cot²A)+(1+tan²A) = 7+tan²A+cot²A = RHS
Proved.
(ix)
LHS = (cosecA−sinA)(secA−cosA) = (1sinA−sinA)(1cosA−cosA) = [1−sin²AsinA][1−cos²AcosA]
= (cos²AsinA)(sin²AcosA) = sinA·cosA
RHS = 1tanA+cotA = 1/(sinAcosA+cosAsinA) = 1/[sin²A+cos²AsinAcosA] = sinAcosA1 = sinAcosA
= (cos²AsinA)(sin²AcosA) = sinA·cosA
RHS = 1tanA+cotA = 1/(sinAcosA+cosAsinA) = 1/[sin²A+cos²AsinAcosA] = sinAcosA1 = sinAcosA
LHS = RHS = sinA cosA. Proved.
(x)
First part: 1+tan²A1+cot²A = sec²Acosec²A = (1cos²A)×sin²A = tan²A (as in Q3(iv))
Second part: [1−tanA1−cotA]². Write cotA=1tanA:
(1−tanA)/(1−1tanA) = (1−tanA)/[tanA−1tanA] = tanA1−tanAtanA−1 = tanA·[−(tanA−1)]/(tanA−1) = −tanA
So [1−tanA1−cotA]² = (−tanA)² = tan²A
Second part: [1−tanA1−cotA]². Write cotA=1tanA:
(1−tanA)/(1−1tanA) = (1−tanA)/[tanA−1tanA] = tanA1−tanAtanA−1 = tanA·[−(tanA−1)]/(tanA−1) = −tanA
So [1−tanA1−cotA]² = (−tanA)² = tan²A
Both expressions equal tan²A. Proved.
Frequently Asked Questions
Any two right triangles with the same acute angle are similar (AA similarity — the right angle and the shared acute angle fix the third angle too). Similar triangles have sides in the same ratio, and a trigonometric ratio is exactly a ratio of two sides, so it stays fixed however large or small the triangle is.
Treat the given ratio as a fraction of two sides (say sin A = k1 means opposite:hypotenuse = k:1), assign those two sides using a common multiplier, find the third side with the Pythagoras theorem, then read every ratio directly off the completed triangle — see Exercise 8.1, Q3–Q5.
Pick the side with more terms (usually the LHS) and simplify it down to match the other side, rather than working both sides separately. Converting everything to sin and cos first often reveals a common factor or a difference-of-squares pattern that collapses the expression quickly — used throughout Exercise 8.3, Q4.
tan A = sin A / cos A, and cos 90° = 0. Dividing by zero is undefined, so tan 90° (and sec 90°, which is 1cos 90°) simply doesn't exist as a number — the same reasoning makes cot 0° and cosec 0° undefined, since sin 0° = 0.
Yes — a printable Formula Card covering all 7 units, including Introduction to Trigonometry, is available for ₹99.
Continue to the Next Chapter
Chapter 9 (Some Applications of Trigonometry) puts these ratios to work on heights and distances.