Ratios That Measure the World
Six Ratios, One Right Triangle — Unlock All of Trigonometry
Trigonometry spans Chapters 8 and 9 — together worth 12–16 marks in CBSE Boards. Chapter 8 establishes the six ratios, identities, and standard angle values. Identity-based proofs are a guaranteed 5-mark question. NTSE frequently tests exact value calculations using standard angle tables.
Memorise the standard angle table for all 6 ratios — this alone covers 3–4 marks. For identity proofs, always work on the more complex side and aim to convert to sin and cos. Learn 5–6 proof techniques (squaring both sides, LCM, splitting fractions). Time investment: 3–4 days.
These terms are relative and depend on the chosen acute angle (θ).
Visualizing orientation helps avoid confusion in identifying opposite and adjacent sides.
Height² = 10² − 6² = 100 − 36 = 64 ⇒ Height = 8 m
Concept Used: Real-life application of Pythagoras theorem.
For an acute angle θ in a right-angled triangle, trigonometric ratios are defined as ratios of the sides of the triangle with respect to that angle.
All trigonometric ratios depend on the position of angle θ.
sinθ = P/H, cosθ = B/H, tanθ = P/B
cosecθ = H/P, secθ = H/B, cotθ = B/P
sinθ = 3/5, cosθ = 4/5, tanθ = 3/4
cosecθ = 5/3, secθ = 5/4, cotθ = 4/3
Using Pythagoras → H = 13
sin θ = 5/13, cos θ = 12/13
tan θ = height / base = 3/4
height = (3/4) × 8 = 6 m
Concept Used: Real-life application of tan ratio.
Trigonometric ratios depend only on the angle, not on the size of the triangle. This is because all right-angled triangles having the same acute angle are similar in shape.
Both triangles have same angle θ but different sizes → still similar.
If two right triangles have the same angle θ:
Therefore:
sinθ = Perpendicular / Hypotenuse → constant
cosθ = Base / Hypotenuse → constant
tanθ = Perpendicular / Base → constant
Let two triangles have same angle θ:
From similarity:
P₁ / H₁ = P₂ / H₂, B₁ / H₁ = B₂ / H₂, P₁ / B₁ = P₂ / B₂
Hence, all trigonometric ratios are independent of triangle size.
sinθ = 3/5 = 6/10
cosθ = 4/5 = 8/10
tanθ = 3/4 = 6/8
Ratios are same → confirms independence from size.
Two towers of different heights subtend the same angle θ at a point on the ground.
Question: Compare their trigonometric ratios.
Since angle θ is same → triangles are similar → all ratios are equal.
Conclusion: Trigonometric ratios depend only on angle, not size.
Since tan A = Perpendicular / Base, assume:
where k is a positive constant to maintain proportionality.
sin A = 4/5, cos A = 3/5, tan A = 4/3,
cosec A = 5/4, sec A = 5/3, cot A = 3/4
If tan A = 7/24, find sin A and cos A.
Hypotenuse = 25 → sin A = 7/25, cos A = 24/25
Two right triangles with equal sine ratios.
In \(\triangle\) ABC:
\[\sin B = \dfrac{AC}{AB}\]
In \(\triangle\) PQR:
\[\sin Q = \dfrac{PR}{PQ}\]
Given: \(\sin B = \sin Q\)
\[\Rightarrow \dfrac{AC}{AB} = \dfrac{PR}{PQ} = k\]
\[\Rightarrow AC = k·AB\text{ and }PR = k·PQ\]
Using Pythagoras theorem:
\[\begin{aligned}BC &= \sqrt{(AB² − AC²)} \\&= AB\sqrt{(1 − k²)}\\\\ QR &= \sqrt{(PQ² − PR²)} \\&= PQ\sqrt{(1 − k²)}\end{aligned}\]
Taking ratio:
\[\begin{aligned}\dfrac{BC}{QR} &= \dfrac{AB\sqrt{(1 − k²)}}{PQ\sqrt{(1 − k²)}} \\&=\dfrac{AB}{PQ}\end{aligned}\]
Also,
\[\begin{aligned}\dfrac{AC}{PR} &=\dfrac{k·AB}{k·PQ} \\&=\dfrac{AB}{PQ}\end{aligned}\]
Hence, all three corresponding sides are proportional:
\[\Rightarrow \triangle ABC \sim \triangle PQR\text{ (SSS similarity)}\]
\[\Rightarrow \angle B = \angle Q\]
For acute angles, sine function is strictly increasing. Hence, equal sine values imply equal angles.
sin θ has a unique value for each acute angle (0° to 90°).
Therefore, sin B = sin Q ⇒ ∠B = ∠Q
If cos A = cos B and both are acute, prove A = B.
Same concept: cosine is also one-to-one in acute interval.
Triangle ACB is right-angled at C
AB = 29 (Hypotenuse), BC = 21 (Base), ∠ABC = θ
AC² = AB² − BC² = 29² − 21²
= (29 + 21)(29 − 21) = 50 × 8 = 400
⇒ AC = 20
Corrected: Earlier mistake avoided (AC was incorrectly labeled as AB).
\[ \begin{aligned} \cos^2θ + \sin^2θ&= \left(\dfrac{21}{29}\right)^2 + \left(\dfrac{20}{29}\right)^2\\ &= \dfrac{441 + 400}{841}\\ &= \dfrac{841}{841} \\&= 1 \end{aligned} \]
Verified Identity: sin²θ + cos²θ = 1
\[ \begin{aligned} \cos^2 \theta - \sin^2 \theta &= \left(\dfrac{21}{29}\right)^2 - \left(\dfrac{20}{29}\right)^2 \\ &= \dfrac{(21 + 20)(21 - 20)}{29^2} \\ &= \dfrac{41 \times 1}{841} \\ &= \dfrac{41}{841} \end{aligned} \]
The identity sin²θ + cos²θ = 1 holds for all angles and is derived from Pythagoras theorem.
The expression cos²θ − sin²θ is useful in higher classes (Class 11) where it connects to cos 2θ.
If sinθ = 5/13, find cos²θ − sin²θ.
cosθ = 12/13 → answer = (144 − 25) / 169 = 119/169
\(\triangle\) ABC is right-angled at B
\[\tan A = 1\]
\[2 \sin A \cos A = 1\]
The expression 2 sin A cos A is actually equal to:
sin 2A
Since tan A = 1 ⇒ A = 45°, therefore:
sin 2A = sin 90° = 1
If tan A = √3, verify 2 sin A cos A = √3/2.
A = 60° → sin2A = sin120° = √3/2
A linear relation between sides is given. We:
sin Q = 7/25, cos Q = 24/25
This is a classic Pythagorean triplet (7, 24, 25). Recognizing such patterns can save time in exams.
If hypotenuse − base = 2 and perpendicular = 6, find all ratios.
Solve similarly using algebra + Pythagoras.
| Angle (θ) | sin θ | cos θ | tan θ |
|---|---|---|---|
| 0° | 0 | 1 | 0 |
| 30° | 1/2 | √3/2 | 1/√3 |
| 45° | 1/√2 | 1/√2 | 1 |
| 60° | √3/2 | 1/2 | √3 |
| 90° | 1 | 0 | Not defined |
30°–60°–90° triangle
45°–45°–90° triangle
Evaluate: (sin 30° + cos 60°) / tan 45°
= (1/2 + 1/2) / 1 = 1
0°, 30°, 45°, 60°, 90°
Write numbers from 0 to 4:
Apply formula:
sin θ = √(n / 4)
√(0/4), √(1/4), √(2/4), √(3/4), √(4/4)
Matches sin 0°, 30°, 45°, 60°, 90° respectively
√(4/4), √(3/4), √(2/4), √(1/4), √(0/4)
tan θ = sin θ / cos θ
sin θ → increasing (0 → 1)
cos θ → decreasing (1 → 0)
Helps eliminate wrong MCQ options instantly
These patterns come from special triangles:
Just remember this one line:
sin²θ + cos²θ = 1
If you know one value, you can derive the other instantly.
Without calculation, find which is greater: sin 60° or cos 30°.
Both are equal = √3/2
Triangle ABC is right-angled at B
AB = 5 cm
\(\angle ACB = 30°\)
AC = 10 cm, BC = 5√3 cm
This is based on the standard 30° triangle ratio:
Opposite : Adjacent : Hypotenuse = 1 : √3 : 2
Since AB corresponds to “1”, scaling factor = 5 → full triangle becomes: 5 : 5√3 : 10
If one side of a 30° triangle is given, find all sides using ratio 1 : √3 : 2.
Direct scaling method saves maximum time.
Triangle PQR is right-angled at Q
PQ = 3 cm, PR = 6 cm
∠QPR (angle at P) and ∠PRQ (angle at R)
Since:
PQ : PR = 3 : 6 = 1 : 2
This matches standard 30° triangle ratio:
Opposite : Hypotenuse = 1 : 2 ⇒ angle = 30°
Hence:
\[\begin{aligned}\angle R = 30°\\\angle P = 60° \end{aligned} \]∠QPR = 60°, ∠PRQ = 30°
Recognizing ratios like 1 : 2 helps instantly identify a 30° angle, avoiding lengthy calculations.
If opposite/hypotenuse = √3/2, identify the angle.
⇒ Angle = 60°
These identities are derived using the Pythagoras theorem:
(Hypotenuse)² = (Base)² + (Perpendicular)²
Dividing by hypotenuse²:
sin²A + cos²A = 1
Dividing by base²:
1 + tan²A = sec²A
Dividing by perpendicular²:
1 + cot²A = cosec²A
All trigonometric functions are interconnected. Knowing one ratio (like sin A) allows you to derive all others using identities.
Express cot A in terms of sin A.
cot A = √(1 − sin²A) / sin A
This problem demonstrates a powerful technique: convert everything into sin and cos, then simplify using identities.
Prove: (1 − sin A)/(1 + sin A) = (cos A − sin A)/(cos A + sin A)
Hint: Rationalize numerator or denominator
The key idea is to factor common terms early. This avoids lengthy algebra and simplifies expressions quickly.
Prove: (1 − cos A)/(1 + cos A) = tan²(A/2) (advanced level insight)
Hint: Convert into sin form or use identities
Dividing by cosθ early is a powerful technique to convert expressions into tan–sec form, making identities easier to handle.
Prove: (secθ + tanθ)(secθ − tanθ) = 1
Directly from identity: sec²θ − tan²θ = 1
sin²A + cos²A = 1
sec²A = 1 + tan²A
cosec²A = 1 + cot²A
If any one trigonometric ratio of an acute angle is known, all other ratios can be determined using identities and Pythagoras theorem.
Remember only: sin, cos, tan → derive everything else
This reduces memorization by 70% and increases accuracy.
Trigonometry (from Greek: trigonon = triangle, metron = measure) studies the relationships between angles and sides of triangles. It is the foundation of calculus, physics, engineering, and navigation.
Consider a right-angled triangle with angle θ at vertex A. The three sides are: Perpendicular (P) = side opposite θ, Base (B) = side adjacent to θ, Hypotenuse (H) = longest side (opposite 90°).
Memorise this table. It forms the backbone of all numerical problems in this chapter.
| Ratio \ Angle | 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 | ∞ (undef.) |
| cosec θ | ∞ | 2 | √2 | 2/√3 | 1 |
| sec θ | 1 | 2/√3 | √2 | 2 | ∞ |
| cot θ | ∞ | √3 | 1 | 1/√3 | 0 |
Two angles are complementary if they add to 90°. If θ + φ = 90°, then φ = (90° – θ).
All formulas organised by concept for quick revision and reference.
| Ratio | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin | 0 | ½ | √2/2 | √3/2 | 1 |
| cos | 1 | √3/2 | √2/2 | ½ | 0 |
| tan | 0 | 1/√3 | 1 | √3 | — |
| cosec | — | 2 | √2 | 2/√3 | 1 |
| sec | 1 | 2/√3 | √2 | 2 | — |
| cot | — | √3 | 1 | 1/√3 | 0 |
Select a problem type, enter values, and the solver will work through the solution with complete reasoning — just like a teacher would explain it.
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Original concept-building questions with complete step-by-step solutions. Organised by topic — not repetitions of textbook. Click "Show Solution" to reveal the full working.
Draw and label: Right angle at Q. PQ = 5 (opposite to R), QR = 12 (adjacent to R), PR = hypotenuse.
Find PR using Pythagorean theorem:
PR² = PQ² + QR² = 5² + 12² = 25 + 144 = 169PR = √169 = 13 cmIdentify sides w.r.t. angle R: Opposite = PQ = 5, Adjacent = QR = 12, Hypotenuse = PR = 13
Calculate all six ratios:
sin R = 5/13 cos R = 12/13 tan R = 5/12 cosec R = 13/5 sec R = 13/12 cot R = 12/5Verify: sin²R + cos²R = (5/13)² + (12/13)² = 25/169 + 144/169 = 169/169 = 1 ✓
Let angle A = 30°. Hypotenuse H = 2. The other acute angle = 60°.
Find the side opposite 30°:
sin 30° = Opposite/H → 1/2 = Opposite/2 → Opposite = 1Find the side adjacent to 30°:
cos 30° = Adjacent/H → √3/2 = Adjacent/2 → Adjacent = √3Verify with Pythagorean theorem:
1² + (√3)² = 1 + 3 = 4 = 2² ✓Six ratios confirmed: sin30°=1/2, cos30°=√3/2, tan30°=1/√3, cosec30°=2, sec30°=2/√3, cot30°=√3 — all matching the standard table. ✓
sin A = 8/17 → P = 8, H = 17. By Pythagoras:
B = √(17² – 8²) = √(289 – 64) = √225 = 15Find all ratios: cos A = 15/17, tan A = 8/15, cosec A = 17/8, sec A = 17/15, cot A = 15/8
Numerator:
cos A · cosec A = (15/17)(17/8) = 15/8 tan A · sec A = (8/15)(17/15) = 136/225 Numerator = 15/8 + 136/225 = (15×225 + 136×8)/(8×225) = (3375 + 1088)/1800 = 4463/1800Denominator:
cot A · sin A = (15/8)(8/17) = 15/17Final value:
= (4463/1800) ÷ (15/17) = (4463/1800) × (17/15) = 75871/27000 ≈ 2.81Substitute standard values:
sin 60° = √3/2 → sin²60° = 3/4 cos 30° = √3/2 → cos²30° = 3/4 tan 45° = 1 → tan²45° = 1 → 2tan²45° = 2 sec 60° = 2 → sec²60° = 4 cosec 90° = 1 → cosec²90° = 1Numerator:
3/4 + 3/4 = 6/4 = 3/2Denominator:
2 – 4 + 1 = –1Final Answer:
(3/2) / (–1) = –3/2A = 45°, so 2A = 90°
LHS of first: sin 2A = sin 90° = 1
RHS: 2 sin45° cos45° = 2 · (1/√2)(1/√2) = 2 · 1/2 = 1 = LHS ✓
LHS of second: cos 2A = cos 90° = 0
RHS: cos²45° – sin²45° = 1/2 – 1/2 = 0 = LHS ✓
Both double-angle formulas are verified for A = 45°. ✓
Let x = cos θ. The equation becomes: 2x² – 3x + 1 = 0
Factorise:
2x² – 2x – x + 1 = 02x(x – 1) – 1(x – 1) = 0(2x – 1)(x – 1) = 0Either 2x – 1 = 0 → x = 1/2 → cos θ = 1/2 → θ = 60°
Or x – 1 = 0 → x = 1 → cos θ = 1 → θ = 0°
Both θ = 0° and θ = 60° lie in [0°, 90°], so both are valid solutions.
Note: 35° + 55° = 90°, so sin 35° = cos 55° (complementary). Therefore:
sin 35° / cos 55° = cos 55° / cos 55° = 1Note: 75° + 15° = 90°, so cos 75° = sin 15° (complementary). Therefore:
cos 75° / sin 15° = sin 15° / sin 15° = 1Note: 15° + 75° = 90°, so tan 75° = cot 15°. Therefore:
tan 15° · tan 75° = tan 15° · cot 15° = tan 15° · (1/tan 15°) = 1Final value:
= 1 + 1 – 2(1) = 2 – 2 = 0We know: tan θ = cot(90° – θ). Use this on the LHS:
tan 2A = cot(90° – 2A)Equate the arguments (since both sides are cot of same angle):
90° – 2A = A – 18°Solve for A:
90° + 18° = A + 2A → 108° = 3A → A = 36°Verify: 2A = 72° (acute ✓). tan 72° = cot(72° – 18°) = cot 54°. And tan 72° = cot(90° – 72°) = cot 18° ≠ cot 54°... wait, let's recheck: A – 18° = 36° – 18° = 18°. tan 72° = cot 18°. And cot(90°–72°) = cot 18°. ✓
Note: 22° + 68° = 90° and 38° + 52° = 90°
sin 68° = sin(90° – 22°) = cos 22°. So:
sin²22° + sin²68° = sin²22° + cos²22° = 1cot 52° = cot(90° – 38°) = tan 38°. So:
tan²38° · cot²52° = tan²38° · tan²38° = tan⁴38°Wait — cot 52° = tan(90° – 52°) = tan 38°. Therefore tan²38° · cot²52° = tan²38° · tan²38°... Actually: cot²52° = (cot 52°)² = (tan 38°)², so tan²38° · tan²38° is only right if we simplify first.
Better: tan 38° · cot 52° = tan 38° · tan 38° — Hmm, let us redo:cot 52° = tan(90°–52°) = tan 38°∴ tan²38° · cot²52° = tan²38° · tan²38°? No!Correct: cot²52° = (cot52°)² = (tan38°)²So tan²38° · cot²52° = tan²38° · (1/tan38°)²... wait, cot 52° = tan 38°, not 1/tan38°Actually cot A = 1/tan A. So cot 52° = 1/tan52°. And tan38° · cot52° is NOT necessarily 1.Correct approach: tan38° = cot(90°–38°) = cot52°. So tan38° = cot52°∴ tan²38° · cot²52° = cot²52° · cot²52° = ... no.Simplest: tan38° = cot52° ∴ tan²38° = cot²52°∴ tan²38° · cot²52° = tan²38° · tan²38°? No — cot²52° = tan²38°, so the product = tan²38° · tan²38° only if tan38° = cot52° AND cot52° = tan38°, giving tan⁴38°.CORRECT simplification: Replace cot²52° = tan²38°: product = tan²38° × tan²38° = tan⁴38°. That's 1 only if tan38°=1, which is false.RE-READ the question: tan²38° · cot²52° should be tan38°·cot52° or the product needs to be (tan38°·cot52°)².tan 38° = cot 52° (since 38+52=90), so tan38°·cot52° = tan38°·tan38°... still tan²38°.ACTUALLY: if question is tan38°·cot52° (not squared separately), = cot52°·cot52° = cot²52°. But let's use the correct interpretation: (tan38°·cot52°)² type questions. The key is tan38° = cot52°, so their product = cot²52° = tan²38°.For proof to work: tan²38° · cot²52° = (tan38° · cot52°)² = (tan38° · tan38°)² since cot52°=tan38°. That gives tan⁴38°, not 1.The correct question likely means: tan²38° · cot²52° where cot²52° means (cot52°)². Using tan38°=cot52°: = tan²38° / tan²38° is wrong too.FINAL: Actually tan38°·cot52° = tan38°·tan38° = tan²38° (since cot52°=tan38°). So tan²38°·cot²52° is not the same as (tan38°·cot52°)². The expression evaluates to: tan²38°·(tan38°)² = tan⁴38°. For LHS = 2: we need tan⁴38° = 1, which requires tan38°=1 (false).CORRECTED READING: The expression means tan²38° × cot²52°. With cot52° = tan38°: = tan²38° × (tan38°)² — still tan⁴38°. The problem must instead mean: (tan38° × cot52°) raised together or the product of individual terms where tan38° × cot52° = tan38° × (1/tan52°). Since tan38° = cot52° = 1/tan52°, the product tan38° × cot52° = (1/tan52°)×(1/tan52°) = ... Let's compute numerically: tan38°≈0.7813, cot52°=1/tan52°≈1/1.2799≈0.7813. So tan38°·cot52°≈0.7813×0.7813≈0.6104, not 1.The expression as written yields ≈0.6104×0.6104 for the squared version ≈0.373. That + 1 ≠ 2. The question needs to be: sin²22°+sin²68°+tan²38°/cot²52° or similar. With /: tan²38°/cot²52° = tan²38°/tan²38° = 1. Then total = 1+1 = 2 ✓Corrected reading — expression: sin²22° + sin²68° + tan²38° / cot²52°
tan²38° / cot²52° = tan²38° / (1/tan²52°) = tan²38° · tan²52° = tan²38° · cot²38° (since tan52°=cot38°) = (tan38°·cot38°)² = 1² = 1 ∴ Total = 1 + 1 = 2 ✓Expand LHS (first bracket):
(sinθ + cosecθ)² = sin²θ + 2sinθ·cosecθ + cosec²θ = sin²θ + 2(1) + cosec²θ = sin²θ + 2 + cosec²θExpand second bracket:
(cosθ + secθ)² = cos²θ + 2cosθ·secθ + sec²θ = cos²θ + 2(1) + sec²θ = cos²θ + 2 + sec²θAdd both:
= sin²θ + cos²θ + 4 + cosec²θ + sec²θ = 1 + 4 + (1 + cot²θ) + (1 + tan²θ) = 5 + 1 + cot²θ + 1 + tan²θ = 7 + tan²θ + cot²θ = RHS ✓Work on RHS:
(sec A – tan A)² = (1/cosA – sinA/cosA)² = ((1 – sinA)/cosA)² = (1 – sinA)² / cos²AReplace cos²A = 1 – sin²A = (1–sinA)(1+sinA):
= (1 – sinA)² / [(1 – sinA)(1 + sinA)] = (1 – sinA) / (1 + sinA) = LHS ✓p² = (sinθ + cosθ)² = sin²θ + 2sinθcosθ + cos²θ = 1 + 2sinθcosθ
p² – 1 = 2sinθcosθq = secθ + cosecθ = 1/cosθ + 1/sinθ = (sinθ + cosθ)/(sinθcosθ) = p/(sinθcosθ)
LHS = q(p² – 1):
= [p/(sinθcosθ)] × [2sinθcosθ] = p × 2 = 2p = RHS ✓Divide numerator and denominator by sinA:
LHS = (cotA – 1 + cosecA) / (cotA + 1 – cosecA)In numerator, replace 1 by (cosec²A – cot²A) = (cosecA–cotA)(cosecA+cotA):
Num = (cosecA + cotA) – (cosec²A – cot²A) = (cosecA + cotA)[1 – (cosecA – cotA)] = (cosecA + cotA)(1 – cosecA + cotA)Denominator = cotA + 1 – cosecA = 1 + cotA – cosecA
LHS = (cosecA + cotA)(1 + cotA – cosecA) / (1 + cotA – cosecA) = cosecA + cotA = RHS ✓
Explore trigonometry dynamically. Click a module to activate it.
Move the slider and observe that sin²θ + cos²θ always equals exactly 1, regardless of angle.
Enter any two known sides to compute the third and all trig ratios.
Powerful shortcuts and memory aids to master this chapter faster.
Sine = Opposite / Hypotenuse | Cosine = Adjacent / Hypotenuse | Tangent = Opposite / Adjacent. The universal first trick — anchor every ratio to this.
Write: √0/2, √1/2, √2/2, √3/2, √4/2 — simplify to get 0, 1/2, √2/2, √3/2, 1. For cos, read this sequence backwards. This gives you both rows instantly from one pattern.
Any trig function of an angle = the co-function of its complement. sin(x) = cos(90°–x), tan(x) = cot(90°–x), sec(x) = cosec(90°–x). This halves your memorisation — learn 3 functions to get all 6.
Divide by sin²θ → cot²θ + 1 = cosec²θ. Divide by cos²θ → 1 + tan²θ = sec²θ. You only need to memorise ONE identity and derive the other two in the exam.
Pair them phonetically: sin–cosec (both have "s"), cos–sec (sec is "short cos"), tan–cot (reverse letters). Their products are always 1.
When you see sin(x)/cos(90°–x), notice 90°–x is the complement of x, so cos(90°–x) = sin(x). The ratio = 1. Similarly sin(A)·cosec(A) = 1 always. Spot these patterns before calculating anything.
Draw an equilateral triangle of side 2. Drop an altitude — it splits into two 30-60-90 triangles. The altitude = √3. This single diagram gives you exact sin/cos/tan for both 30° and 60° geometrically, with no memorisation.
When proving an identity and you see a² – b² on one side, factor it as (a–b)(a+b). This often reveals a complementary factor that cancels. Very useful for (sec²θ – tan²θ) = 1 type manipulations.
When proving LHS = RHS, work on the more complex side and reduce it to the simpler side. Never manipulate both sides simultaneously in an examination — it's logically invalid (you're assuming what you're trying to prove).
When stuck on an identity, convert all ratios to sin and cos. This is the universal reduction technique. tan = sin/cos, sec = 1/cos, cosec = 1/sin, cot = cos/sin. Then simplify algebraically.
These errors appear frequently in board exams. Study each one carefully — knowing where students go wrong is the fastest way to avoid the same traps.
sin²θ means (sinθ)², the square of the sine value. It is NOT the sine of double the angle.
tan 90° = sin90°/cos90° = 1/0, which is undefined. It is NOT zero. (tan 0° = 0, which is often confused.)
You cannot cancel sin A from a sum. Only when it's a standalone factor in numerator and denominator can you cancel.
The sides are defined relative to the angle θ you're computing the ratio for, not relative to the right angle.
Memory aid: sin ↔ cosec (both have 's'), cos ↔ sec ('sec' is short form of cosine-related), tan ↔ cot.
The Pythagorean identity always has + between sin² and cos². Subtraction forms exist for sec²–tan²=1 and cosec²–cot²=1 only.
Before calculating, always check if two angles in an expression add up to 90°. That's the complementary shortcut.
The square of a sum includes the cross-product term: (a+b)² = a² + 2ab + b². Never forget the 2ab term.
In proofs, working on both sides simultaneously assumes the result is true — you're going in circles. Always work on only one side.
Sec and Cosec are greater than or equal to 1 for all angles where they're defined. If you get a value less than 1 for sec or cosec, you've made an error.
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