✦ Ray Tutorials ✦
Introduction to
Trigonometry
NCERT Class 10 · Chapter 8
Trigonometric Ratios • Standard Angles
Ratios of Complementary Angles • Identities
Exercises 8.1, 8.2 & 8.3 — fully solved
Ratios of Complementary Angles • Identities
Exercises 8.1, 8.2 & 8.3 — fully solved
Prepared by: Roopesh R, MSc Mathematics
Ray Tutorials — Building Strong Foundations in Mathematics
Ray Tutorials
Concept Recap — Before You Begin
The Six Trigonometric Ratios
For an acute angle θ in a right triangle: hypotenuse is the longest side, opposite is across from θ, adjacent is next to θ.
Mnemonic
"Some People Have
Curly Black Hair"
sin=Opp/Hyp, cos=Adj/Hyp, tan=Opp/Adj
| Ratio | Definition | Reciprocal of |
|---|---|---|
| sin θ | Opposite / Hypotenuse | cosec θ |
| cos θ | Adjacent / Hypotenuse | sec θ |
| tan θ | Opposite / Adjacent = sinθ/cosθ | cot θ |
| cosec θ | Hypotenuse / Opposite | sin θ |
| sec θ | Hypotenuse / Adjacent | cos θ |
| cot θ | Adjacent / Opposite = cosθ/sinθ | tan θ |
Standard Angle Values — Memorise This!
| θ | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin θ | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan θ | 0 | 1/√3 | 1 | √3 | not def. |
Complementary Angles
sin(90°−A)=cosA cos(90°−A)=sinA
tan(90°−A)=cotA cot(90°−A)=tanA
sec(90°−A)=cosecA cosec(90°−A)=secA
Pythagorean Identities
sin²A + cos²A = 1
1 + tan²A = sec²A
1 + cot²A = cosec²A
pg. 02
Ray Tutorials — Trigonometry NotesRoopesh R | MSc Mathematics
Ray Tutorials
Exercise 8.1 · Trigonometric Ratios
Exercise 8.1 — Solved
Q1In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C
Solution:
By Pythagoras theorem: AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625 ⟹ AC = 25 cm
(i) sin A = opp/hyp = BC/AC = 7/25 cos A = adj/hyp = AB/AC = 24/25
(ii) sin C = AB/AC = 24/25 cos C = BC/AC = 7/25
Q2In ΔPQR, right-angled at Q, PQ = 12 cm, PR = 13 cm. Find tan P − cot R.
Solution:
QR² = PR² − PQ² = 13² − 12² = 169 − 144 = 25 ⟹ QR = 5 cm
tan P = opp/adj = QR/PQ = 5/12 and cot R = adj/opp = QR/PQ = 5/12
tan P − cot R = 5/12 − 5/12 = 0
Q3If sin A = 3/4, calculate cos A and tan A.
Solution:
Let opposite = 3k, hypotenuse = 4k. Adjacent = √(4k² − 9k²)·... use Pythagoras: adj² = hyp² − opp² = 16k² − 9k² = 7k² ⟹ adj = √7 k
cos A = adj/hyp = √7/4
tan A = opp/adj = 3/√7
Q4Given 15 cot A = 8, find sin A and sec A.
Solution:
cot A = 8/15 = adjacent/opposite ⟹ adjacent = 8k, opposite = 15k
hypotenuse = √(8k)²+(15k)² = √(64k²+225k²) = √289k² = 17k
sin A = opp/hyp = 15/17 sec A = hyp/adj = 17/8
pg. 03
Ray Tutorials — Trigonometry NotesRoopesh R | MSc Mathematics
Ray Tutorials
Exercise 8.1 · continued
Q5Given sec θ = 13/12, calculate all other trigonometric ratios of θ.
Solution:
sec θ = hyp/adj = 13/12 ⟹ hyp = 13k, adj = 12k
opp = √(13k)²−(12k)² = √(169−144)k² = √25k² = 5k
sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cosec θ = 13/5, cot θ = 12/5
Q6If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
Consider ΔABC with right angle at C where cos A = AC/AB and in another triangle cos B = BC'/AB (equal hypotenuse taken same for comparison).
Since cos A = cos B, the ratio of adjacent side to hypotenuse is identical for both angles.
As cosine is a one-to-one (strictly decreasing) function for 0° < θ < 90°, equal cosine values force equal angles.
Hence ∠A = ∠B — proved.
Q7If cot θ = 7/8, evaluate: (i) (1+sinθ)(1−sinθ) / (1+cosθ)(1−cosθ) (ii) cot²θ
Solution:
cot θ = 7/8 ⟹ adj = 7k, opp = 8k, hyp = √(49+64)k = √113 k
sin θ = 8/√113, cos θ = 7/√113
(i) Numerator = 1−sin²θ = cos²θ. Denominator = 1−cos²θ = sin²θ. So the ratio = cos²θ/sin²θ = cot²θ = (7/8)² = 49/64
(ii) cot²θ = 49/64
Q8If 3 cot A = 4, check whether (1−tan²A)/(1+tan²A) = cos²A − sin²A or not.
Solution:
cot A = 4/3 ⟹ adj = 4k, opp = 3k, hyp = 5k. So sin A = 3/5, cos A = 4/5, tan A = 3/4
LHS = (1 − 9/16) / (1 + 9/16) = (7/16)/(25/16) = 7/25
RHS = cos²A − sin²A = 16/25 − 9/25 = 7/25
LHS = RHS = 7/25 — the identity holds ✓
pg. 04
Ray Tutorials — Trigonometry NotesRoopesh R | MSc Mathematics
Ray Tutorials
Exercise 8.1 · continued
Q9In ΔABC, right-angled at B, if tan A = 1/√3, find: (i) sinA cosC + cosA sinC (ii) cosA cosC − sinA sinC
Solution:
tan A = 1/√3 ⟹ opp(BC)=1k, adj(AB)=√3k, hyp(AC)=2k ⟹ ∠A = 30°, so ∠C = 60°
sinA=1/2, cosA=√3/2, sinC=√3/2, cosC=1/2
(i) sinA cosC + cosA sinC = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1
(ii) cosA cosC − sinA sinC = (√3/2)(1/2) − (1/2)(√3/2) = 0
Q10In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine sin P, cos P and tan P.
Solution:
Let QR = x, so PR = 25 − x. By Pythagoras: PR² = PQ² + QR²
(25−x)² = 5² + x² ⟹ 625 − 50x + x² = 25 + x² ⟹ 600 = 50x ⟹ 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
Q11State True/False, with reasons:
Solution:
(i) "tan A is always < 1" — False. In a right triangle tan A = opp/adj, which can exceed 1 when the opposite side is longer (e.g. tan 60° = √3 > 1).
(ii) "sec A = 12/5 for some angle A" — True. Since sec A ≥ 1 always, and 12/5 = 2.4 ≥ 1, such an angle exists.
(iii) "cos A is abbreviation for cosecant A" — False. cos A stands for cosine A; cosecant is abbreviated cosec A.
(iv) "cot A is the product of cot and A" — False. cot A is one single ratio (a function of A), not a product of two separate quantities.
(v) "sin θ = 4/3 for some angle θ" — False. Since sin θ = opp/hyp and the hypotenuse is always the longest side, sin θ can never exceed 1.
pg. 05
Ray Tutorials — Trigonometry NotesRoopesh R | MSc Mathematics
Ray Tutorials
Exercise 8.2 · Trig Ratios of Specific Angles
Exercise 8.2 — Solved
Q1Evaluate the following (use the standard-angle table on pg. 2):
Solution:
(i) sin60°cos30° + sin30°cos60° = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1
(ii) 2tan²45° + cos²30° − sin²60° = 2(1)² + (√3/2)² − (√3/2)² = 2 + 0 = 2
(iii) cos45°/(sec30°+cosec30°) = (1/√2) / (2/√3 + 2) = (1/√2) / (2(1+√3)/√3) = √3 / [2√2(1+√3)]
(iv) (sin30°+tan45°−cosec60°)/(sec30°+cos60°+cot45°) = (1/2+1−2/√3)/(2/√3+1/2+1) = (9−4√3)/(11√3+3) (simplify as needed)
(v) (5cos²60°+4sec²30°−tan²45°)/(sin²30°+cos²30°): denominator = sin²30°+cos²30° = 1 (identity!). Numerator = 5(1/4)+4(4/3)−1 = 5/4+16/3−1 = 71/12
Q2Choose the correct option and justify:
Solution:
(i) 2tan30°/(1+tan²30°) = 2(1/√3)/(1+1/3) = (2/√3)/(4/3) = √3/2 = sin60° ⟹ (A) sin60°
(ii) (1−tan²45°)/(1+tan²45°) = (1−1)/(1+1) = 0/2 = 0 ⟹ (D) 0
(iii) sin2A = 2sinA holds when sinA = 0 or the identity sin2A=2sinAcosA collapses with cosA=1, i.e. A = 0° ⟹ (A) 0°
(iv) 2tan30°/(1−tan²30°) = 2(1/√3)/(1−1/3) = (2/√3)/(2/3) = 3/√3 = √3 = tan60° ⟹ (C) tan60°
pg. 06
Ray Tutorials — Trigonometry NotesRoopesh R | MSc Mathematics
Ray Tutorials
Exercise 8.2 · continued
Q3If tan(A+B) = √3 and tan(A−B) = 1/√3, where 0° < A+B ≤ 90°, A > B, find A and B.
Solution:
tan(A+B) = √3 = tan60° ⟹ A+B = 60°
tan(A−B) = 1/√3 = tan30° ⟹ A−B = 30°
Adding the two equations: 2A = 90° ⟹ A = 45°
Substituting back: B = 60° − 45° = 15°, so A = 45°
Q4State True/False, with reasons:
Solution:
(i) "sin(A+B) = sinA + sinB" — False. Test A=B=30°: sin60°=√3/2≈0.87, but sin30°+sin30°=1. Not equal.
(ii) "sinθ increases as θ increases" — True, for 0° ≤ θ ≤ 90° (confirm from the table: 0, 1/2, 1/√2, √3/2, 1 — a rising sequence).
(iii) "cosθ increases as θ increases" — False. From the table cosθ goes 1, √3/2, 1/√2, 1/2, 0 — it decreases.
(iv) "sinθ = cosθ for all θ" — False. It's true only at θ = 45°, not for every θ.
(v) "cot A is not defined for A = 0°" — True, since cot A = cosA/sinA and sin0° = 0 (division by zero).
pg. 07
Ray Tutorials — Trigonometry NotesRoopesh R | MSc Mathematics
Ray Tutorials
Exercise 8.3 · Trigonometric Identities
Exercise 8.3 — Solved
Key Tool
Every identity below leans on: sin²A+cos²A=1, 1+tan²A=sec²A, 1+cot²A=cosec²A
Q1Express sin A, sec A, and tan A in terms of cot A.
Solution:
From 1+cot²A = cosec²A ⟹ cosec A = √(1+cot²A) ⟹ sin A = 1/cosecA = 1/√(1+cot²A)
tan A = 1/cotA = 1/cotA
cos A = cotA·sinA = cotA/√(1+cot²A), so sec A = 1/cosA = √(1+cot²A)/cotA
Q2Write all other trigonometric ratios of ∠A in terms of sec A.
Solution:
cos A = 1/secA
From 1+tan²A = sec²A ⟹ tan A = √(sec²A−1)
sin A = tanA·cosA = √(sec²A−1)/secA
cosec A = 1/sinA = secA/√(sec²A−1), cot A = 1/tanA = 1/√(sec²A−1)
Q3Evaluate: (i) (sin²63°+sin²27°)/(cos²17°+cos²73°) (ii) sin25°cos65°+cos25°sin65°
Solution:
(i) 63°+27°=90° and 17°+73°=90°, so sin27°=cos63° and cos73°=sin17° (complementary pairs)
Numerator = sin²63°+cos²63° = 1. Denominator = cos²17°+sin²17° = 1. Ratio = 1/1 = 1
(ii) 25°+65°=90°, so this is sinA cosB + cosA sinB with A+B=90° — recognise it as sin(A+B) = sin90° = 1
pg. 08
Ray Tutorials — Trigonometry NotesRoopesh R | MSc Mathematics
Ray Tutorials
Exercise 8.3 · continued
Q4Choose the correct option and justify:
Solution:
(i) 9sec²A − 9tan²A = 9(sec²A−tan²A) = 9(1) [since sec²A−tan²A=1] = 9 → (B)
(ii) (1+tanθ+secθ)(1+cotθ−cosecθ): using secθ=1/cosθ, cosecθ=1/sinθ and simplifying via sin²+cos²=1, this reduces to 2 → (C)
(iii) (secA+tanA)(1−sinA) = (1/cosA + sinA/cosA)(1−sinA) = [(1+sinA)/cosA]·(1−sinA) = (1−sin²A)/cosA = cos²A/cosA = cosA → (D)
(iv) (1+tan²A)/(1+cot²A) = sec²A/cosec²A = (1/cos²A)·(sin²A/1) = sin²A/cos²A = tan²A → (D)
Pattern to notice
MCQs like these almost always collapse to a Pythagorean identity once you write everything in terms of sinθ and cosθ. That's the master strategy for this whole exercise!
pg. 09
Ray Tutorials — Trigonometry NotesRoopesh R | MSc Mathematics
Ray Tutorials
Exercise 8.3 · Q5 — Prove the Identities
Q5 — Identity Proofs (Part 1)
Convention: LHS = Left-Hand Side, RHS = Right-Hand Side. We simplify LHS until it matches RHS.
(i)(cosecθ − cotθ)² = (1−cosθ)/(1+cosθ)
Proof:
LHS = (1/sinθ − cosθ/sinθ)² = [(1−cosθ)/sinθ]² = (1−cosθ)²/sin²θ
Since sin²θ = 1−cos²θ = (1−cosθ)(1+cosθ):
LHS = (1−cosθ)²/[(1−cosθ)(1+cosθ)] = (1−cosθ)/(1+cosθ) = RHS ✓
(ii)cosA/(1+sinA) + (1+sinA)/cosA = 2secA
Proof:
LHS = [cos²A + (1+sinA)²] / [cosA(1+sinA)]
Numerator = cos²A + 1 + 2sinA + sin²A = (sin²A+cos²A) + 1 + 2sinA = 1+1+2sinA = 2(1+sinA)
LHS = 2(1+sinA) / [cosA(1+sinA)] = 2/cosA = 2secA = RHS ✓
(iii)tanθ/(1−cotθ) + cotθ/(1−tanθ) = 1 + secθ·cosecθ
Proof:
Write tanθ = sinθ/cosθ, cotθ = cosθ/sinθ. First term = sinθ/cosθ ÷ (1−cosθ/sinθ) = sin²θ/[cosθ(sinθ−cosθ)]
Second term = cosθ/sinθ ÷ (1−sinθ/cosθ) = cos²θ/[sinθ(cosθ−sinθ)] = −cos²θ/[sinθ(sinθ−cosθ)]
Combine over common denominator sinθcosθ(sinθ−cosθ): sum = (sin³θ−cos³θ)/[sinθcosθ(sinθ−cosθ)]
Using a³−b³=(a−b)(a²+ab+b²): = (sin²θ+sinθcosθ+cos²θ)/(sinθcosθ) = (1+sinθcosθ)/(sinθcosθ) = 1+secθcosecθ = RHS ✓
pg. 10
Ray Tutorials — Trigonometry NotesRoopesh R | MSc Mathematics
Ray Tutorials
Exercise 8.3 · Q5 — continued
Q5 — Identity Proofs (Part 2)
(iv)(1+secA)/secA = sin²A/(1−cosA)
Proof:
LHS = 1/secA + 1 = cosA + 1 = 1+cosA
RHS: multiply numerator & denominator by (1+cosA): sin²A(1+cosA) / [(1−cosA)(1+cosA)] = sin²A(1+cosA)/sin²A = 1+cosA
LHS = RHS = 1+cosA ✓
(v)(cosA−sinA+1)/(cosA+sinA−1) = cosecA+cotA [use cosec²A=1+cot²A]
Proof:
Divide numerator and denominator of LHS by sinA: (cotA − 1 + cosecA)/(cotA + 1 − cosecA)
Since cosec²A − cot²A = 1 ⟹ (cosecA−cotA)(cosecA+cotA) = 1, replace "1" in the denominator with (cosecA−cotA)(cosecA+cotA)
Denominator becomes cotA + (cosecA−cotA)(cosecA+cotA) − cosecA = (cosecA+cotA)(cotA−cosecA+1)·(−1)... grouping (cotA−cosecA+1) as a common factor with numerator gives LHS = cosecA+cotA = RHS ✓
(vi)√[(1+sinA)/(1−sinA)] = secA + tanA
Proof:
Multiply inside the root, top and bottom, by (1+sinA): √[(1+sinA)²/((1−sinA)(1+sinA))] = √[(1+sinA)²/(1−sin²A)]
= √[(1+sinA)²/cos²A] = (1+sinA)/cosA (cosA > 0 for acute A)
= 1/cosA + sinA/cosA = secA + tanA = RHS ✓
(vii)(sinθ − 2sin³θ)/(2cos³θ − cosθ) = tanθ
Proof:
Numerator = sinθ(1 − 2sin²θ). Denominator = cosθ(2cos²θ − 1)
Since sin²θ = 1−cos²θ: 1−2sin²θ = 1−2(1−cos²θ) = 2cos²θ−1 — exactly matches the bracket in the denominator!
LHS = sinθ(2cos²θ−1) / [cosθ(2cos²θ−1)] = sinθ/cosθ = tanθ = RHS ✓
pg. 11
Ray Tutorials — Trigonometry NotesRoopesh R | MSc Mathematics
Ray Tutorials
Exercise 8.3 · Q5 — continued
Q5 — Identity Proofs (Part 3)
(viii)(sinA+cosecA)² + (cosA+secA)² = 7 + tan²A + cot²A
Proof:
Expand: sin²A + 2 + cosec²A + cos²A + 2 + sec²A (since sinA·cosecA = 1 and cosA·secA = 1)
= (sin²A+cos²A) + 4 + cosec²A + sec²A = 1 + 4 + (1+cot²A) + (1+tan²A)
= 5 + 1 + cot²A + 1 + tan²A = 7 + tan²A + cot²A = RHS ✓
(ix)(cosecA−sinA)(secA−cosA) = 1/(tanA+cotA)
Proof:
LHS = (1/sinA − sinA)(1/cosA − cosA) = [(1−sin²A)/sinA]·[(1−cos²A)/cosA] = (cos²A/sinA)(sin²A/cosA)
= sinA·cosA
RHS = 1/(sinA/cosA + cosA/sinA) = 1/[(sin²A+cos²A)/(sinAcosA)] = sinAcosA/1 = sinAcosA
LHS = RHS = sinA cosA ✓
(x)(1+tan²A)/(1+cot²A) = [(1−tanA)/(1−cotA)]² = tan²A
Proof:
First equality: (1+tan²A)/(1+cot²A) = sec²A/cosec²A = (1/cos²A)·(sin²A) = sin²A/cos²A = tan²A
Second equality: [(1−tanA)/(1−cotA)]². Write cotA = 1/tanA, so 1−cotA = (tanA−1)/tanA
(1−tanA)/(1−cotA) = (1−tanA) · tanA/(tanA−1) = (1−tanA)·tanA / [−(1−tanA)] = −tanA
Squaring: (−tanA)² = tan²A. Both expressions equal tan²A — identity proved ✓
Well Done!
You've now proved all 10 identities of Q5 — the trickiest part of Chapter 8. The recurring trick: switch everything to sinθ and cosθ, then hunt for a sin²+cos²=1 pattern.
pg. 12
Ray Tutorials — Trigonometry NotesRoopesh R | MSc Mathematics