Class 12 Maths · Chapter 2
Short answer:
Is chapter me 2 exercises hain — Exercise 2.1 (14 questions, principal values) aur Miscellaneous Exercise (17 questions, properties + proofs). Yahan 18 representative questions solve kiye gaye hain jo principal value nikalna, sin−1x/cos−1x/tan−1x ki properties use karke simplify karna, aur inverse trig identities prove karna cover karte hain.
Inverse trigonometric functions trig functions ko 'reverse' karte hain — par sirf ek restricted domain (principal value branch) par, kyunki trig functions many-one hote hain. Jaise sinx ek se zyada x ke liye same value deta hai, isliye sin−1x define karne ke liye range ko [−π/2, π/2] tak restrict karna padta hai taaki ek unique answer mile. Ye chapter har inverse trig function ka domain-range define karta hai, unki properties sikhata hai, aur inhe expressions simplify karne me use karta hai — jo integration (Class 12 hi me aage) ke liye foundation banata hai.
Chapter 2 Summary — 5 Minute Revision
1. Kyun principal value branch chahiye
Trigonometric functions (sinx, cosx, tanx, etc.) many-one hote hain — infinite x values same output dete hain (jaise sin(π/6) = sin(5π/6) = 1/2). Ek function tabhi invertible hota hai jab woh bijective (one-one + onto) ho. Isliye inverse trig function define karne ke liye domain ko restrict karke ek branch choose ki jaati hai jaha function one-one aur onto ho jaaye — isko principal value branch kehte hain.
2. Domain aur Range (Principal Value Branch) — har function ka
| Function | Domain | Range (Principal Value Branch) |
|---|---|---|
| sin−1x | [−1, 1] | [−π/2, π/2] |
| cos−1x | [−1, 1] | [0, π] |
| tan−1x | R (all reals) | (−π/2, π/2) |
| cot−1x | R (all reals) | (0, π) |
| sec−1x | R − (−1, 1) | [0, π] − {π/2} |
| cosec−1x | R − (−1, 1) | [−π/2, π/2] − {0} |
↔ Table ko side me swipe karein
3. Graphs (in words)
- y = sin−1x: increasing curve, x-axis pe [−1,1] se y-axis pe [−π/2, π/2] tak jaata hai — origin se symmetric (odd function), y = sinx ke graph ka mirror image line y=x ke around (restricted domain me).
- y = cos−1x: decreasing curve, [−1,1] → [0, π]. x=0 par y=π/2, x=1 par y=0, x=−1 par y=π.
- y = tan−1x: increasing, S-shaped curve, saari real line par defined, do horizontal asymptotes y = π/2 aur y = −π/2 ke beech bounded rehta hai.
- y = cot−1x: decreasing curve, (0, π) me bounded, asymptotes y=0 aur y=π.
- sec−1x aur cosec−1x ke graphs disjoint branches me hote hain kyunki domain me (−1,1) excluded hai.
4. Key Properties (do groups)
Group 1 — Reciprocal relations:
- sin−1(1/x) = cosec−1x, for |x| ≥ 1
- cos−1(1/x) = sec−1x, for |x| ≥ 1
- tan−1(1/x) = cot−1x, for x > 0
Group 2 — Odd/even nature:
- sin−1(−x) = −sin−1x
- tan−1(−x) = −tan−1x
- cosec−1(−x) = −cosec−1x
- cos−1(−x) = π − cos−1x
- sec−1(−x) = π − sec−1x
- cot−1(−x) = π − cot−1x
Group 3 — Complementary pairs (sum = π/2):
- sin−1x + cos−1x = π/2, for x ∈ [−1,1]
- tan−1x + cot−1x = π/2, for x ∈ R
- sec−1x + cosec−1x = π/2, for |x| ≥ 1
Group 4 — Sum/difference formulas (tan−1):
- tan−1x + tan−1y = tan−1[(x+y)/(1−xy)], jab xy < 1
- tan−1x − tan−1y = tan−1[(x−y)/(1+xy)], jab xy > −1
- 2tan−1x = tan−1[2x/(1−x²)], jab |x| < 1
5. sin−1(sinx) type simplification ka approach
sin−1(sinx) = x sirf tab jab x ∈ [−π/2, π/2]. Agar x is range se bahar hai, toh pehle x ko [−π/2, π/2] ke andar equivalent angle me convert karo (using periodicity/symmetry of sine), phir formula apply karo. Same logic cos−1(cosx) = x ke liye [0, π] range ke saath.
Poore Class 12 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–Q18)
Q1. Find the principal value of sin−1(−1/2).
Hume woh angle θ chahiye jaha sinθ = −1/2 aur θ ∈ [−π/2, π/2].
sin(−π/6) = −1/2, aur −π/6 ∈ [−π/2, π/2]
Isliye sin−1(−1/2) = −π/6.
Q2. Find the principal value of cos−1(−1/2).
Hume θ chahiye jaha cosθ = −1/2 aur θ ∈ [0, π].
cos(2π/3) = −1/2, aur 2π/3 ∈ [0, π]
Isliye cos−1(−1/2) = 2π/3.
Q3. Find the principal value of cosec−1(2).
Hume θ chahiye jaha cosecθ = 2, i.e. sinθ = 1/2, aur θ ∈ [−π/2, π/2] − {0}.
sin(π/6) = 1/2
Isliye cosec−1(2) = π/6.
Q4. Find the principal value of tan−1(−√3).
Hume θ chahiye jaha tanθ = −√3 aur θ ∈ (−π/2, π/2).
tan(π/3) = √3 ⟹ tan(−π/3) = −√3
Isliye tan−1(−√3) = −π/3.
Q5. Find the principal value of cos−1(−1/√2).
Hume θ chahiye jaha cosθ = −1/√2 aur θ ∈ [0, π].
cos(3π/4) = −1/√2
Isliye cos−1(−1/√2) = 3π/4.
Q6. Find the principal value of tan−1(−1).
Hume θ chahiye jaha tanθ = −1 aur θ ∈ (−π/2, π/2).
tan(π/4) = 1 ⟹ tan(−π/4) = −1
Isliye tan−1(−1) = −π/4.
Q7. Find the principal value of sec−1(2/√3).
Hume θ chahiye jaha secθ = 2/√3, i.e. cosθ = √3/2, aur θ ∈ [0, π] − {π/2}.
cos(π/6) = √3/2
Isliye sec−1(2/√3) = π/6.
Q8. Evaluate tan−1(1) + cos−1(−1/2) + sin−1(−1/2).
Har term ki principal value nikalo:
tan−1(1) = π/4
cos−1(−1/2) = 2π/3
sin−1(−1/2) = −π/6
Sum:
π/4 + 2π/3 − π/6 = 3π/12 + 8π/12 − 2π/12 = 9π/12 = 3π/4
Isliye answer = 3π/4.
Q9. Find the value of tan−1(1) + cos−1(1/2) + sin−1(1/2).
Har term ki principal value:
tan−1(1) = π/4, cos−1(1/2) = π/3, sin−1(1/2) = π/6
Sum:
π/4 + π/3 + π/6 = 3π/12 + 4π/12 + 2π/12 = 9π/12 = 3π/4
Isliye answer = 3π/4.
Q10. Find the value of cos−1(1/2) + 2sin−1(1/2).
Values substitute karo:
cos−1(1/2) = π/3, sin−1(1/2) = π/6
π/3 + 2(π/6) = π/3 + π/3 = 2π/3
Isliye answer = 2π/3.
Q11. Prove that sin−1(2x√(1−x²)) = 2sin−1x, for −1/√2 ≤ x ≤ 1/√2.
Let x = sinθ, jaha θ = sin−1x aur θ ∈ [−π/4, π/4] (given restriction se).
2x√(1−x²) = 2sinθ·cosθ = sin(2θ)
Kyunki θ ∈ [−π/4, π/4], isliye 2θ ∈ [−π/2, π/2] — yahi sin−1 ki principal range hai, toh:
sin−1(2x√(1−x²)) = sin−1(sin2θ) = 2θ = 2sin−1x
Hence proved.
Q12. Express tan−1(cosx / (1 − sinx)), −π/2 < x < π/2, in the simplest form.
Half-angle substitution: cosx = cos²(x/2) − sin²(x/2), 1 − sinx = (cos(x/2) − sin(x/2))².
cosx = (cos(x/2) − sin(x/2))(cos(x/2) + sin(x/2))
cosx/(1−sinx) = (cos(x/2) + sin(x/2)) / (cos(x/2) − sin(x/2))
Numerator-denominator ko cos(x/2) se divide karo:
= (1 + tan(x/2)) / (1 − tan(x/2)) = tan(π/4 + x/2)
Isliye tan−1[cosx/(1−sinx)] = π/4 + x/2.
Q13. Write tan−1[√(1−cosx)/(1+cosx)], 0 ≤ x < π, in the simplest form.
Half-angle identities use karo:
1 − cosx = 2sin²(x/2), 1 + cosx = 2cos²(x/2)
√[(1−cosx)/(1+cosx)] = √[tan²(x/2)] = |tan(x/2)| = tan(x/2), kyunki 0 ≤ x/2 < π/2
Isliye tan−1[tan(x/2)] = x/2.
Q14. Solve for x: tan−1[(1−x)/(1+x)] = (1/2)tan−1x, x > 0.
LHS ko rewrite karo. Note that (1−x)/(1+x) form tan(π/4 − θ) jaisi hai jaha x = tanθ:
tan−1[(1−x)/(1+x)] = tan−1(1) − tan−1x = π/4 − tan−1x (valid since x>0)
Given equation:
π/4 − tan−1x = (1/2)tan−1x
π/4 = (3/2)tan−1x ⟹ tan−1x = π/6
x = tan(π/6) = 1/√3
Isliye x = 1/√3.
Q15. Find the value of tan−1√3 − sec−1(−2).
Values nikalo:
tan−1√3 = π/3
sec−1(−2): secθ = −2 ⟹ cosθ = −1/2, θ ∈ [0,π]−{π/2} ⟹ θ = 2π/3
Difference:
π/3 − 2π/3 = −π/3
Isliye answer = −π/3.
Q16. Prove: 2tan−1(1/2) + tan−1(1/7) = tan−1(31/17).
Pehle 2tan−1(1/2) ko double-angle formula se simplify karo:
2tan−1(1/2) = tan−1[2(1/2)/(1−(1/2)²)] = tan−1[1/(3/4)] = tan−1(4/3)
Ab tan−1(4/3) + tan−1(1/7) ko addition formula se combine karo (xy = 4/21 < 1, so formula directly applicable):
= tan−1[(4/3 + 1/7)/(1 − (4/3)(1/7))] = tan−1[(28/21 + 3/21)/(1 − 4/21)]
= tan−1[(31/21)/(17/21)] = tan−1(31/17)
Hence proved, LHS = RHS.
Q17. Prove that cos−1(4/5) + cos−1(12/13) = cos−1(33/65).
Let A = cos−1(4/5), so cosA = 4/5, sinA = 3/5 (A in first quadrant since 4/5 > 0).
Let B = cos−1(12/13), so cosB = 12/13, sinB = 5/13.
cos(A+B) = cosA·cosB − sinA·sinB = (4/5)(12/13) − (3/5)(5/13)
= 48/65 − 15/65 = 33/65
Kyunki A, B dono [0, π/2] me hain, A+B ∈ [0, π], toh:
A + B = cos−1(33/65)
Hence proved.
Q18. Find the value of sin[π/3 − sin−1(−1/2)].
Pehle andar ki value nikalo:
sin−1(−1/2) = −π/6
Substitute karo:
sin[π/3 − (−π/6)] = sin[π/3 + π/6] = sin(π/2) = 1
Isliye answer = 1.
Important Equations — Ek Nazar Me
| Function | Domain | Range (Principal Value Branch) |
|---|---|---|
| sin−1x | [−1, 1] | [−π/2, π/2] |
| cos−1x | [−1, 1] | [0, π] |
| tan−1x | R | (−π/2, π/2) |
| cot−1x | R | (0, π) |
| sec−1x | R − (−1, 1) | [0, π] − {π/2} |
| cosec−1x | R − (−1, 1) | [−π/2, π/2] − {0} |
↔ Table ko side me swipe karein
| Identity | Condition |
|---|---|
| sin−1(1/x) = cosec−1x | |x| ≥ 1 |
| cos−1(1/x) = sec−1x | |x| ≥ 1 |
| tan−1(1/x) = cot−1x | x > 0 |
| sin−1(−x) = −sin−1x | x ∈ [−1,1] |
| tan−1(−x) = −tan−1x | x ∈ R |
| cosec−1(−x) = −cosec−1x | |x| ≥ 1 |
| cos−1(−x) = π − cos−1x | x ∈ [−1,1] |
| sec−1(−x) = π − sec−1x | |x| ≥ 1 |
| cot−1(−x) = π − cot−1x | x ∈ R |
| sin−1x + cos−1x = π/2 | x ∈ [−1,1] |
| tan−1x + cot−1x = π/2 | x ∈ R |
| sec−1x + cosec−1x = π/2 | |x| ≥ 1 |
| tan−1x + tan−1y = tan−1[(x+y)/(1−xy)] | xy < 1 |
| tan−1x − tan−1y = tan−1[(x−y)/(1+xy)] | xy > −1 |
| 2tan−1x = tan−1[2x/(1−x²)] | |x| < 1 |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Range ke bahar answer dena. sin−1(−1/2) ka answer 7π/6 ya 11π/6 likhna galat hai kyunki ye [−π/2, π/2] range se bahar hai — sahi answer −π/6 hai. Hamesha check karo answer principal value branch ke andar hai.
- tan−1x + tan−1y formula me xy<1 condition bhoolna. Agar xy > 1 (dono x,y positive), toh formula me π add/subtract karna padta hai — directly tan−1[(x+y)/(1−xy)] likhna wrong answer dega.
- sin−1x ko (sinx)−1 = cosecx samajhna. sin−1x ek alag function hai (inverse function), reciprocal nahi. Notation confusion se marks katte hain.
- Negative argument pe sign error. cos−1(−x) = π − cos−1x hota hai, −cos−1x nahi (jabki sin−1(−x) = −sin−1x hota hai). Odd aur even-type behavior ko function ke hisaab se yaad rakho, sabko same formula mat samjho.
- sin−1(sinx) = x hamesha samajhna. Ye sirf tab true hai jab x ∈ [−π/2, π/2]. Agar x is range se bahar hai (jaise x = 5π/6), toh pehle equivalent angle range ke andar nikalna padega.
- Domain check kiye bina expression simplify karna. Jaise Q11/Q12/Q13 jaisi identities sirf ek specific x-range ke liye valid hoti hain — proof likhte waqt given restriction (jaise −1/√2 ≤ x ≤ 1/√2) explicitly mention na karna step marks katwa sakta hai.
Board-Style Important Questions
- 1 mark: Find the principal value of cos−1(−1/√2).
- 1 mark: Write the domain of the function sec−1x.
- 2 marks: Find the value of tan−1(1) + tan−1(1/2) + tan−1(1/3).
- 4 marks: Prove that tan−1[√(1+x) − √(1−x)] / [√(1+x) + √(1−x)] = π/4 − (1/2)cos−1x, for −1/√2 ≤ x ≤ 1.
- 4 marks: Show that sin−1(3/5) − sin−1(8/17) = cos−1(84/85).
- 2 marks: Express cot−1(1/√(x²−1)) in the simplest form, for x > 1.
Aksar Poochhe Jaane Wale Sawaal
Inverse trig functions ka principal value branch kya hota hai?
Trig functions many-one hote hain, toh unka inverse define karne ke liye domain restrict karke ek branch choose karte hain jaha function one-one aur onto ho — usko principal value branch kehte hain. Jaise sin−1x ke liye ye [−π/2, π/2] hai.
sin−1x aur cosecx me kya difference hai?
sin−1x ek inverse function hai jo output angle deta hai, jabki cosecx = 1/sinx reciprocal function hai. In dono ko confuse mat karo — notation similar dikhta hai par matlab bilkul alag hai.
sin−1(sinx) hamesha x ke barabar hota hai kya?
Nahi, sirf tab jab x already [−π/2, π/2] range me ho. Agar x is range se bahar hai, toh pehle equivalent angle isi range ke andar nikalna padega.
tan−1x + tan−1y formula kab directly apply hoti hai?
Jab xy < 1 ho. Agar xy > 1 aur x,y dono positive hain, toh answer me π add karna padta hai; agar x,y dono negative hain toh π subtract karna padta hai.
cos−1x aur sin−1x ka range same kyun nahi hai?
Har function ka apna principal value branch hota hai jo ek convention se decide kiya gaya hai — cos ke liye [0,π] isliye chosen hai kyunki wahan cos strictly decreasing aur one-one hai, jabki sin ke liye [−π/2,π/2] me sin strictly increasing hai.
Kya sec−1x aur cosec−1x ka domain (−1,1) include karta hai?
Nahi, in dono ka domain R − (−1,1) hai, matlab −1 aur 1 ke beech ki values (excluding boundary) domain me nahi aati kyunki secx aur cosecx ki range hamesha |value| ≥ 1 hoti hai.
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