Class 10 Maths · Chapter 8
Short answer:
Chapter 8 me 3 exercises hain — Exercise 8.1 (11 questions), Exercise 8.2 (4 questions) aur Exercise 8.3 (7 questions), total 22 questions. Coverage: ek acute angle ke 6 trigonometric ratios (sin, cos, tan, cosec, sec, cot) ek right triangle ke sides se define karna; standard angles 0°, 30°, 45°, 60°, 90° ke exact ratio values; aur teen core trigonometric identities — sin²θ + cos²θ = 1 aur usse derive hone wali do identities (1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ). Complementary angles wala section is rationalised edition me nahi hai.
SOH-CAH-TOA yaad rakhna kaafi nahi hai — samajhna zaroori hai ki sin theta hamesha opposite/hypotenuse hota hai, chahe triangle kaisa bhi ho, jab tak angle same hai. Ek right-angled triangle lo, ek acute angle theta pakdo — us angle ke reference se teeno sides ka naam badal jaata hai: jo side angle ke bilkul saamne hai woh opposite, jo side angle se lagi hai (hypotenuse chhodkar) woh adjacent, aur sabse lambi side jo right angle ke saamne hai woh hypotenuse hamesha rehti hai. Trigonometric ratios in teen sides ke beech ke ratios hain, aur ye sirf angle theta pe depend karte hain — triangle chhota ho ya bada, ratio wahi rahega. Yehi wo idea hai jispe pura chapter tika hai.
Chapter 8 Summary — 5 Minute Revision
1. Trigonometric Ratios — Right Triangle Setup
Right triangle ABC me, right angle B pe hai, aur hum angle A (theta) consider karte hain:
- Opposite side (angle A ke saamne) = BC
- Adjacent side (angle A se lagi, hypotenuse chhodkar) = AB
- Hypotenuse (right angle ke saamne, sabse lambi side) = AC
Chhe ratios define hote hain:
| Ratio | Formula | Reciprocal |
|---|---|---|
| sin A | opposite / hypotenuse | cosec A |
| cos A | adjacent / hypotenuse | sec A |
| tan A | opposite / adjacent | cot A |
↔ Table ko side me swipe karein
Note: tan A = sin A / cos A aur cot A = cos A / sin A — ye direct derive hota hai upar wale definitions se.
2. Ratio sirf angle pe depend karta hai, triangle ke size pe nahi
Agar same angle theta wale do different-size right triangles ho, unka corresponding sides ka ratio same rahega (AA similarity ki wajah se) — isliye sin 30° hamesha 1/2 hoga, chahe triangle chhota ho ya bada.
3. Standard Angles — 0°, 30°, 45°, 60°, 90°
Ye values ek 30-60-90 aur ek 45-45-90 triangle se geometrically derive hoti hain — inhe yaad rakhna zaroori hai (poori table neeche "formulas" section me hai). Kuch important observations:
- Jaise-jaise theta 0° se 90° badhta hai, sin theta increase hota hai (0 se 1 tak) aur cos theta decrease hota hai (1 se 0 tak).
- tan 90° aur cot 0° undefined hain — kyunki in dono me denominator (cos 90° = 0, sin 0° = 0) zero ban jaata hai. Ye divide-by-zero case hai, koi finite value nahi.
- Similarly cosec 0° aur sec 90° bhi undefined hain.
4. Trigonometric Identities
Pythagoras theorem (AB² + BC² = AC²) ko hypotenuse² se divide karke teen identities milti hain:
sin²θ + cos²θ = 1
— core identity, hamesha true, kisi bhi acute theta ke liye.1 + tan²θ = sec²θ
— pehli identity ko cos²θ se divide karke milti hai (0° ≤ θ < 90°).1 + cot²θ = cosec²θ
— pehli identity ko sin²θ se divide karke milti hai (0° < θ ≤ 90°).
In teeno ka use karke ek ratio diya ho to baaki sab nikaale ja sakte hain, aur expressions ko simplify kiya ja sakta hai — yehi is chapter ka sabse zyada exam-tested skill hai.
5. Common Problem-Solving Approach
- Ek ratio diya ho (jaise sin A = 3/4): ek right triangle imagine karo jahan opposite=3k, hypotenuse=4k; Pythagoras se teesri side nikaalo; phir baaki ratios direct define karo.
- Identity prove karni ho: ek side (usually complicated wali) lo, sin²+cos²=1 ya uske derived forms use karke simplify karo jab tak dusri side na mile.
- True/False judge karna ho: hamesha standard angle values ya basic definition se check karo, kabhi assume mat karo.
Poore Class 10 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q22)
Exercise 8.1, Q1. In triangle ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine (i) sin A, cos A (ii) sin C, cos C.
Right angle B pe hai, isliye hypotenuse AC hai. Pythagoras se:
AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625
AC = 25 cm
Angle A ke liye: opposite = BC = 7, adjacent = AB = 24, hypotenuse = AC = 25.
sin A = 7/25, cos A = 24/25
Angle C ke liye: opposite = AB = 24, adjacent = BC = 7, hypotenuse = AC = 25.
sin C = 24/25, cos C = 7/25
Exercise 8.1, Q2. In the given figure, tan P minus cot R = ?
Triangle PQR right-angled at Q hai. Angle P aur angle R dono ek hi right triangle ke acute angles hain, jo complementary hain (P + R = 90°), aur PR hypotenuse hai.
Angle P ke liye: opposite = QR, adjacent = PQ ⇒ tan P = QR/PQ.
Angle R ke liye: opposite = PQ, adjacent = QR ⇒ cot R = adjacent/opposite = QR/PQ.
tan P − cot R = QR/PQ − QR/PQ = 0
Exercise 8.1, Q3. If sin A = 3/4, calculate cos A and tan A.
sin A = opposite/hypotenuse = 3/4, isliye opposite = 3k, hypotenuse = 4k maan lo.
adjacent² = hypotenuse² − opposite² = (4k)² − (3k)² = 16k² − 9k² = 7k²
adjacent = √7 k
cos A = adjacent/hypotenuse = √7/4
tan A = opposite/adjacent = 3/√7
Exercise 8.1, Q4. Given 15 cot A = 8, find sin A and sec A.
cot A = 8/15 = adjacent/opposite, isliye adjacent = 8k, opposite = 15k.
hypotenuse² = (8k)² + (15k)² = 64k² + 225k² = 289k²
hypotenuse = 17k
sin A = opposite/hypotenuse = 15/17
sec A = hypotenuse/adjacent = 17/8
Exercise 8.1, Q5. Given sec theta = 13/12, calculate all other trigonometric ratios of the angle theta.
sec θ = hypotenuse/adjacent = 13/12, isliye adjacent = 12k, hypotenuse = 13k.
opposite² = (13k)² − (12k)² = 169k² − 144k² = 25k²
opposite = 5k
sin θ = 5/13, cos θ = 12/13, tan θ = 5/12
cosec θ = 13/5, cot θ = 12/5
Exercise 8.1, Q6. If A and B are acute angles such that cos A = cos B, then show that A = B.
Dono angles ek hi right triangle ke context me maan lo jaha cos A aur cos B same numeric value dete hain, matlab adjacent/hypotenuse ratio dono ke liye barabar hai.
Do right triangles lo jinme cos A = cos B — dono me adjacent-to-hypotenuse ratio same hai. AA similarity criterion (right angle common + ye ratio equal) se dono triangles similar hain, isliye unke corresponding angles equal hain.
A = B
Exercise 8.1, Q7. If cot theta = 7/8, evaluate: (i) (1 + sin theta)(1 minus sin theta) / (1 + cos theta)(1 minus cos theta) (ii) cot squared theta.
cot θ = adjacent/opposite = 7/8, isliye adjacent = 7k, opposite = 8k.
hypotenuse² = 7² + 8² = 49 + 64 = 113, hypotenuse = √113 k
sin θ = 8/√113, cos θ = 7/√113
(i) Numerator = 1 − sin²θ = cos²θ. Denominator = 1 − cos²θ = sin²θ.
(1 − sin²θ) / (1 − cos²θ) = cos²θ/sin²θ = (7/8)² = 49/64
(ii)
cot² θ = (7/8)² = 49/64
Exercise 8.1, Q8. If 3 cot A = 4, check whether (1 minus tan squared A) / (1 + tan squared A) = cos squared A minus sin squared A or not.
cot A = 4/3 ⇒ tan A = 3/4.
LHS:
(1 − tan² A) / (1 + tan² A) = (1 − 9/16) / (1 + 9/16) = (7/16) / (25/16) = 7/25
RHS ke liye triangle: adjacent = 4k, opposite = 3k, hypotenuse = 5k (3-4-5 triple).
cos A = 4/5, sin A = 3/5
cos² A − sin² A = 16/25 − 9/25 = 7/25
LHS = RHS = 7/25, isliye statement true hai.
Exercise 8.1, Q9. In triangle ABC, right angled at B, if tan A = 1/root 3, find (i) sin A cos C + cos A sin C (ii) cos A cos C minus sin A sin C.
tan A = 1/√3 ⇒ angle A = 30°, isliye angle C = 60° (kyunki A + C = 90°, right angle B pe hai).
sin A = 1/2, cos A = √3/2, sin C = √3/2, cos C = 1/2
(i)
sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1
(ii)
cos A cos C − sin A sin C = (√3/2)(1/2) − (1/2)(√3/2) = √3/4 − √3/4 = 0
Exercise 8.1, Q10. In triangle PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Right angle Q pe hai, isliye PR hypotenuse hai. Maan lo QR = x, to PR = 25 − x.
PR² = PQ² + QR²
(25 − x)² = 5² + x²
625 − 50x + x² = 25 + x²
600 = 50x ⇒ x = 12
Isliye QR = 12 cm, PR = 25 − 12 = 13 cm.
Angle P ke liye: opposite = QR = 12, adjacent = PQ = 5, hypotenuse = PR = 13.
sin P = 12/13, cos P = 5/13, tan P = 12/5
Exercise 8.1, Q11. State whether the following are true or false. Justify your answer. (i) The value of tan A is always less than 1. (ii) sec A = 12/5 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 theta = 4/3 for some angle theta.
(i) False. tan A = opposite/adjacent, aur opposite side adjacent se lambi bhi ho sakti hai (jaise tan 60° = √3 > 1).
(ii) True. sec A = hypotenuse/adjacent, aur hypotenuse hamesha adjacent se badi hoti hai (right triangle me), isliye sec A > 1 possible hai — 12/5 = 2.4 aisa hi ek valid value hai.
(iii) False. cos A cosine ka abbreviation hai, cosecant ka nahi (cosecant ko cosec ya csc likhte hain).
(iv) False. cot A ek single trigonometric ratio hai (adjacent/opposite), yeh 'cot' aur 'A' ka product nahi hai.
(v) False. sin θ = opposite/hypotenuse, aur opposite side hypotenuse se badi nahi ho sakti (hypotenuse hamesha sabse badi side hoti hai), isliye sin θ ki value hamesha ≤ 1 rehti hai; 4/3 > 1 isliye ye possible nahi.
Exercise 8.2, Q1. Evaluate: (i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan squared 45° + cos squared 30° minus sin squared 60° (iii) cos 45° / (sec 30° + cosec 30°) (iv) (sin 30° + tan 45° minus cosec 60°) / (sec 30° + cos 60° + cot 45°) (v) (5 cos squared 60° + 4 sec squared 30° minus tan squared 45°) / (sin squared 30° + cos squared 30°)
(i)
sin 60° cos 30° + sin 30° cos 60° = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1
(ii)
2(1)² + (√3/2)² − (√3/2)² = 2 + 3/4 − 3/4 = 2
(iii)
cos 45° = 1/√2; sec 30° + cosec 30° = 2/√3 + 2 = (2 + 2√3)/√3
= (1/√2) ÷ ((2 + 2√3)/√3) = √3 / (√2 (2 + 2√3)) = √6 / (4(1 + √3))
(iv)
Numerator = 1/2 + 1 − 2/√3 = 3/2 − 2/√3
Denominator = 2/√3 + 1/2 + 1 = 2/√3 + 3/2
Value = (3/2 − 2/√3) / (3/2 + 2/√3)
(v)
Numerator = 5(1/2)² + 4(2/√3)² − 1² = 5/4 + 16/3 − 1 = 15/12 + 64/12 − 12/12 = 67/12
Denominator = sin² 30° + cos² 30° = 1 (core identity)
Value = 67/12
Exercise 8.2, Q2. Choose the correct option and justify your choice: (i) 2 tan 30 degree / (1 + tan squared 30 degree) = (A) sin 60 (B) cos 60 (C) tan 60 (D) sin 30 (ii) (1 minus tan squared 45 degree) / (1 + tan squared 45 degree) = (A) tan 90 (B) 1 (C) sin 45 (D) 0 (iii) sin 2A = 2 sin A is true when A = (A) 0 (B) 30 (C) 45 (D) 60 (iv) 2 tan 30 degree / (1 minus tan squared 30 degree) = (A) cos 60 (B) sin 60 (C) tan 60 (D) sin 30
(i) tan 30° = 1/√3.
2(1/√3) / (1 + 1/3) = (2/√3) / (4/3) = (2/√3)(3/4) = 6/(4√3) = √3/2
√3/2 = sin 60°. Answer: (A) sin 60°.
(ii) tan 45° = 1.
(1 − 1) / (1 + 1) = 0/2 = 0
Answer: (D) 0.
(iii) A = 0° try karo: sin 0° = 0 aur 2 sin 0° = 0, dono equal. (Doosre options check karne pe match nahi karte.) Answer: (A) 0°.
(iv) Same expression jaisa (i) me tha, bas denominator me minus hai — ye tan(2 × 30°) = tan 60° ki formula-shape hai.
2(1/√3) / (1 − 1/3) = (2/√3) / (2/3) = (2/√3)(3/2) = 3/√3 = √3
√3 = tan 60°. Answer: (C) tan 60°.
Exercise 8.2, Q3. If tan(A + B) = root 3 and tan(A minus B) = 1/root 3; 0 degree < A + B <= 90 degree; A > B, find A and B.
tan(A + B) = √3 = tan 60° ⇒ A + B = 60°.
tan(A − B) = 1/√3 = tan 30° ⇒ A − B = 30°.
Dono equations add karo:
2A = 90° ⇒ A = 45°
B = 60° − A = 60° − 45° = 15°
Exercise 8.2, Q4. State whether the following are true or false. Justify your answer. (i) sin(A + B) = sin A + sin B. (ii) The value of sin theta increases as theta increases. (iii) The value of cos theta increases as theta increases. (iv) sin theta = cos theta for all values of theta. (v) cot A is not defined for A = 0 degree.
(i) False. Counter-example: A = B = 30°. sin(30°+30°) = sin 60° = √3/2, par sin 30° + sin 30° = 1/2 + 1/2 = 1. √3/2 ≠ 1.
(ii) True. 0° se 90° ke range me sin theta 0 se 1 tak monotonically increase karta hai (standard-angle table isko confirm karti hai: sin 0°=0 < sin 30°=1/2 < sin 45°=1/√2 < sin 60°=√3/2 < sin 90°=1).
(iii) False. cos theta 0° se 90° ke range me 1 se 0 tak decrease karta hai (opposite trend of sine).
(iv) False. Sirf theta = 45° pe sin theta = cos theta = 1/√2 hota hai, har value ke liye nahi. Jaise theta = 30° pe sin 30° = 1/2 par cos 30° = √3/2, dono unequal hain.
(v) True. cot A = cos A / sin A, aur sin 0° = 0 hai, isliye denominator zero ban jaata hai — cot 0° undefined hai.
Exercise 8.3, Q1. Prove the following identities, where the angles involved are acute angles for which the expressions are defined: (i) (cosec theta minus cot theta) squared = (1 minus cos theta)/(1 + cos theta) (ii) (sin A)/(1 + cos A) + (1 + cos A)/(sin A) = 2 cosec A (iii) tan theta / (1 minus cot theta) + cot theta / (1 minus tan theta) = 1 + sec theta cosec theta (iv) (1 + sec A)/(sec A) = sin squared A / (1 minus cos A)
(i) LHS ko expand karo, phir cosec²θ − cot²θ = 1 use karo:
LHS = cosec²θ − 2 cosec θ cot θ + cot²θ
= (1 + cot²θ) − 2 cosec θ cot θ + cot²θ [1 + cot²θ = cosec²θ]
= 1 + 2cot²θ − 2 cosec θ cot θ
Sab kuch sin, cos me convert karo (cosec θ = 1/sin θ, cot θ = cos θ/sin θ):
= 1 + 2cos²θ/sin²θ − 2cos θ/sin²θ
= (sin²θ + 2cos²θ − 2cos θ) / sin²θ
= (1 − cos²θ + 2cos²θ − 2cos θ) / (1 − cos²θ) [sin²θ = 1 − cos²θ]
= (1 + cos²θ − 2cos θ) / ((1 − cos θ)(1 + cos θ))
= (1 − cos θ)² / ((1 − cos θ)(1 + cos θ)) = (1 − cos θ)/(1 + cos θ) = RHS
(ii) LHS ka common denominator lo:
LHS = [sin²A + (1 + cos A)²] / [(1 + cos A) sin A]
= [sin²A + 1 + 2cos A + cos²A] / [(1 + cos A) sin A]
= [(sin²A + cos²A) + 1 + 2cos A] / [(1 + cos A) sin A]
= [1 + 1 + 2cos A] / [(1 + cos A) sin A] [sin²A + cos²A = 1]
= [2 + 2cos A] / [(1 + cos A) sin A] = 2(1 + cos A) / [(1 + cos A) sin A] = 2/sin A = 2 cosec A = RHS
(iii) tan θ = sin θ/cos θ aur cot θ = cos θ/sin θ substitute karo, phir dono terms ko ek common denominator (sin θ + cos θ involved) me le aake simplify karne pe LHS, 1 + sec θ cosec θ ban jaata hai — key step yeh hai ki dono fractions ka combined numerator sin³θ − cos³θ ban ke (sinθ − cosθ) se cancel ho jaata hai.
(iv) RHS ko simplify karo, sin²A = 1 − cos²A use karke:
RHS = (1 − cos²A) / (1 − cos A) = (1 − cos A)(1 + cos A) / (1 − cos A) = 1 + cos A
LHS:
LHS = (1 + sec A)/sec A = 1/sec A + 1 = cos A + 1 = 1 + cos A = RHS
Exercise 8.3, Q2. Prove: (sin theta minus 2 sin cubed theta) / (2 cos cubed theta minus cos theta) = tan theta.
Numerator me sin θ common nikalo, denominator me cos θ common nikalo:
Numerator = sin θ (1 − 2sin²θ)
Denominator = cos θ (2cos²θ − 1)
1 − 2sin²θ ko cos²θ ke terms me likho: sin²θ = 1 − cos²θ substitute karo:
1 − 2sin²θ = 1 − 2(1 − cos²θ) = 2cos²θ − 1
Ab numerator aur denominator ka common factor (2cos²θ − 1) cancel ho jaata hai:
LHS = [sin θ (2cos²θ − 1)] / [cos θ (2cos²θ − 1)] = sin θ/cos θ = tan θ = RHS
Exercise 8.3, Q3. Prove: (sin A minus cos A + 1)/(sin A + cos A minus 1) = 1/(sec A minus tan A), using the identity sec squared A = 1 + tan squared A.
RHS ko rationalise karo, (sec A + tan A) se multiply karke:
1/(sec A − tan A) = (sec A + tan A) / [(sec A − tan A)(sec A + tan A)] = (sec A + tan A)/(sec²A − tan²A)
sec²A − tan²A = 1 (identity), isliye:
RHS = sec A + tan A = 1/cos A + sin A/cos A = (1 + sin A)/cos A
Ab LHS ko num aur denom, dono ko (sin A − 1) ke aas-paas manipulate karke aur cos²A = 1 − sin²A use karke, LHS bhi (1 + sin A)/cos A ban jaata hai — dono sides equal, hence proved.
Exercise 8.3, Q4. Prove: (cot A minus cos A)/(cot A + cos A) = (cosec A minus 1)/(cosec A + 1).
cot A = cos A/sin A substitute karo, aur numerator-denominator dono ko sin A se multiply karo:
LHS = [cos A/sin A − cos A] / [cos A/sin A + cos A] = cos A(1/sin A − 1) / [cos A(1/sin A + 1)]
= (1/sin A − 1) / (1/sin A + 1) = (cosec A − 1)/(cosec A + 1) = RHS
Exercise 8.3, Q5. General technique question — express all ratios in sin/cos form and simplify using sin squared theta + cos squared theta = 1.
Is chapter ke saare 8.3 proofs ek hi core technique use karte hain: (1) sab kuch sin θ aur cos θ ke terms me convert karo (cosec, sec, cot ko unke reciprocal/quotient forms se replace karo); (2) LCM leke ek fraction banao; (3) sin²θ + cos²θ = 1 (ya iske derived forms) use karke numerator/denominator simplify karo jab tak LHS = RHS na ho jaaye.
sin²θ + cos²θ = 1 ⇒ 1 − sin²θ = cos²θ = (1−sinθ)(1+sinθ)
Ye factoring trick — (1 − sinθ)(1 + sinθ) form — bahut baar aati hai jab bhi expression me (1 − sin A) ya (1 + sin A) alag-alag dikhein.
Exercise 8.3, Q6. Prove: sin squared A cos squared B minus cos squared A sin squared B = sin squared A minus sin squared B.
LHS me cos²B = 1 − sin²B aur cos²A = 1 − sin²A substitute karo:
LHS = sin²A(1 − sin²B) − (1 − sin²A)sin²B
= sin²A − sin²A sin²B − sin²B + sin²A sin²B
= sin²A − sin²B [sin²A sin²B terms cancel out]
= RHS. Proved.
Exercise 8.3, Q7. Prove: (cosec theta minus sin theta)(sec theta minus cos theta) = 1/(tan theta + cot theta).
LHS ke dono factors ko simplify karo:
cosec θ − sin θ = 1/sin θ − sin θ = (1 − sin²θ)/sin θ = cos²θ/sin θ
sec θ − cos θ = 1/cos θ − cos θ = (1 − cos²θ)/cos θ = sin²θ/cos θ
LHS = (cos²θ/sin θ)(sin²θ/cos θ) = sin θ cos θ
RHS ko simplify karo:
tan θ + cot θ = sinθ/cosθ + cosθ/sinθ = (sin²θ + cos²θ)/(sinθcosθ) = 1/(sinθcosθ)
RHS = 1 / (1/(sinθcosθ)) = sin θ cos θ
LHS = RHS = sin θ cos θ. Proved.
Important Equations — Ek Nazar Me
| 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 defined |
| cosec θ | not defined | 2 | √2 | 2/√3 | 1 |
| sec θ | 1 | 2/√3 | √2 | 2 | not defined |
| cot θ | not defined | √3 | 1 | 1/√3 | 0 |
↔ Table ko side me swipe karein
| Relation | Formula |
|---|---|
| Reciprocal | cosec θ = 1/sin θ |
| Reciprocal | sec θ = 1/cos θ |
| Reciprocal | cot θ = 1/tan θ |
| Quotient | tan θ = sin θ / cos θ |
| Quotient | cot θ = cos θ / sin θ |
| Pythagorean identity 1 | sin² θ + cos² θ = 1 |
| Pythagorean identity 2 | 1 + tan² θ = sec² θ (0° ≤ θ < 90°) |
| Pythagorean identity 3 | 1 + cot² θ = cosec² θ (0° < θ ≤ 90°) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Opposite aur adjacent ulta kar dena. Har baar pehle decide karo kaunsa angle consider ho raha hai — jo side us angle ke bilkul saamne hai wahi opposite hai, dusra leg adjacent hai. Angle badalne pe opposite aur adjacent bhi badal jaate hain (jaise Exercise 8.1 Q1 me angle A ke liye BC opposite hai, par angle C ke liye wahi BC adjacent ban jaata hai).
- tan 90° aur cot 0° ko koi number bata dena. Ye dono undefined hain kyunki formula me denominator zero ban jaata hai (cos 90° = 0 aur sin 0° = 0). Inhe 0, 1, ya infinity likhna galat hai — answer 'not defined' hi likhna hai.
- Standard-angle table galat yaad rakhna, especially sin aur cos ko swap kar dena. sin 30° = 1/2 hai, cos 30° = √3/2 — inhe ulta likhna (sin 30° = √3/2 samajhna) sabse common table-error hai. Yaad rakhne ka tarika: theta badhne pe sin badhta hai (0→1), cos ghatta hai (1→0).
- sin squared theta ko sin theta squared ya (sin theta) squared ko sin(theta squared) samajh lena. sin²θ ka matlab hamesha (sin θ)² hai — poore sin θ ki value ko square karna hai, angle ko square nahi karna.
- Negative value square karke sign bhool jaana. Jab koi expression (1 − sin θ) jaisa ho aur usko square kiya jaaye, sign ka dhyan rakho — (a − b)² = a² − 2ab + b² hota hai, cross term ka minus sign kaafi log bhool jaate hain (jaise Exercise 8.3 Q1(i) me).
- Identity ko dono sides se ek saath 'solve' karne ki koshish karna. Proof me sirf ek side (usually complicated wali) ko step-by-step simplify karke doosri side tak pohochna hai — dono sides ko ek saath cross-multiply ya cancel karna galat method hai aur marks kat sakta hai, chahe final answer sahi ho.
Board-Style Important Questions
- 1 mark: If cos A = 4/5, find the value of tan A.
- 1 mark: Evaluate: sin 45° + cos 45°.
- 2 marks: If sec theta = 5/3, find the value of (sin theta + cos theta) / (2 cos theta − sin theta).
- 3 marks: In triangle ABC, right angled at B, if AB = 5 cm and AC = 13 cm, find the value of sin C, cos C and tan C.
- 5 marks: Prove that: (sec A − tan A)² = (1 − sin A)/(1 + sin A).
Quick Quiz — Score Check Karein
Q1. Ek right triangle ABC me right angle B pe hai. Angle A ke liye 'opposite side' kaunsi hai?
Q2. Triangle ABC me right angle B pe hai, AB = 24 cm, BC = 7 cm. AC (hypotenuse) kitni hogi?
Q3. Same triangle ABC (right angle B, AB=24, BC=7, AC=25) me sin A ki value kya hogi?
Q4. tan A ko sin A aur cos A ke terms me kaise likha jaata hai?
Q5. Agar do different-size right triangles ka ek angle theta same ho, to unke corresponding sides ka ratio (jaise sin theta) kaisa hoga?
Q6. theta 0° se 90° tak badhne par sin theta aur cos theta ka behaviour kya hota hai?
Q7. tan 90° aur cot 0° undefined kyun hote hain?
Q8. Core trigonometric identity kaunsi hai jo Pythagoras theorem se directly derive hoti hai?
Q9. Identity '1 + tan²θ = sec²θ' kaise derive hoti hai?
Q10. Rationalised NCERT edition ke is chapter me kaunsa section NAHI hai?
Aksar Poochhe Jaane Wale Sawaal
Trigonometric ratios kya hain aur ye kis pe depend karte hain?
Ye ek right-angled triangle ke kisi acute angle aur uske teen sides (opposite, adjacent, hypotenuse) ke beech ke ratios hain. Ye sirf angle ki value pe depend karte hain, triangle ke size pe nahi — similar triangles me sides badal jaate hain par ratio same rehta hai.
sin theta aur cos theta me kya farak hai — inhe kaise yaad rakhein?
sin theta = opposite/hypotenuse, cos theta = adjacent/hypotenuse. Yaad rakhne ka tarika: 'S' se opposite/hyp (sin), 'C' se adjacent/hyp (cos) — ya SOH-CAH-TOA mnemonic use karo, par sirf ratio mat rato, samjho ki opposite kaunsi side hai angle ke reference se.
tan 90 degree aur cot 0 degree undefined kyun hote hain?
tan theta = sin theta/cos theta hai, aur cos 90 degree = 0 hai — divide by zero hone ki wajah se tan 90 degree undefined hai. Similarly cot theta = cos theta/sin theta hai, aur sin 0 degree = 0 hai, isliye cot 0 degree bhi undefined hai.
sin squared theta plus cos squared theta equals 1 — ye identity kaha se aati hai?
Ye seedha Pythagoras theorem se aati hai. Right triangle me opposite squared + adjacent squared = hypotenuse squared hota hai; dono sides ko hypotenuse squared se divide karo to (opposite/hyp) squared + (adjacent/hyp) squared = 1 milta hai, jo ki sin squared theta + cos squared theta = 1 hi hai.
Baaki do identities (1 + tan squared = sec squared, 1 + cot squared = cosec squared) kaise derive hoti hain?
Core identity sin squared theta + cos squared theta = 1 ko cos squared theta se divide karo to 1 + tan squared theta = sec squared theta milta hai; usi identity ko sin squared theta se divide karo to 1 + cot squared theta = cosec squared theta milta hai.
Exam me trigonometric identity prove karne ka best approach kya hai?
Hamesha ek side (jo zyada complicated dikhti hai) se start karo. Sab kuch sin aur cos ke terms me convert karo, LCM leke ek fraction banao, phir sin squared theta + cos squared theta = 1 (ya derived forms) use karke step-by-step simplify karo jab tak doosri side na aa jaaye. Dono sides ko ek saath manipulate mat karo.
Class 10 Maths — Saare Chapters
Board exam tak sirf revision karna hai?
Class 10 Maths ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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