Trigonometric Identities šŸ”—

Teen fundamental identities aur unhe use karne ki art ā€” proofs aur examples ke saath.

šŸ”‘ Trigonometric Identity Kya Hai?

šŸ’” Identity vs Equation farq:

Equation sirf kuch specific values par true hoti hai. Jaise x+3=5 sirf x=2 ke liye. Lekin identity sab values ke liye true hoti hai ā€” jaise (a+b)Ā²=aĀ²+2ab+bĀ² hamesha true hai. Trig identities bhi aisa hi hain ā€” kisi bhi angle ke liye valid!

šŸ“ Proof: sinĀ²A + cosĀ²A = 1

Ye sabse important trig identity hai ā€” Pythagoras theorem se directly aati hai.

A B C b (adj) p (opp) h (hyp)
From Pythagoras
pĀ² + bĀ² = hĀ²  ...(in right ā–³ ABC)
Divide both sides by hĀ²
pĀ²/hĀ² + bĀ²/hĀ² = 1
(p/h)Ā² + (b/h)Ā² = 1
By definition
sin A = p/h, cos A = b/h
āˆ“ sinĀ²A + cosĀ²A = 1 ā– 

šŸ”¢ Three Fundamental Identities

Identity 1 ā€” Pythagoras basic
sinĀ²A + cosĀ²A = 1
sinĀ²A = 1 āˆ’ cosĀ²A
cosĀ²A = 1 āˆ’ sinĀ²A
Identity 2 ā€” Divide Identity 1 by cosĀ²A
tanĀ²A + 1 = secĀ²A
tanĀ²A = secĀ²A āˆ’ 1
secĀ²A āˆ’ tanĀ²A = 1 ā†’ (secAāˆ’tanA)(secA+tanA) = 1
Identity 3 ā€” Divide Identity 1 by sinĀ²A
1 + cotĀ²A = cosecĀ²A
cotĀ²A = cosecĀ²A āˆ’ 1
cosecĀ²A āˆ’ cotĀ²A = 1 ā†’ (cosecAāˆ’cotA)(cosecA+cotA) = 1
Factored forms (exam mein bahut kaam aati hain):
(sin+cos)Ā² = 1 + 2sincos
(sināˆ’cos)Ā² = 1 āˆ’ 2sincos
sinā“A + cosā“A = 1 āˆ’ 2sinĀ²AcosĀ²A
(secAāˆ’tanA)(secA+tanA) = 1
(cosecAāˆ’cotA)(cosecA+cotA) = 1

šŸŽÆ Strategy for Proving Identities

5-Step Strategy

  1. LHS ya RHS mein se complex side chunno ā€” wahan se start karo
  2. sin aur cos mein convert karo (sab kuch simplify ho jaata hai)
  3. Algebraic manipulation ā€” common factors, sum/difference of squares
  4. Fundamental identities use karo (sinĀ²+cosĀ²=1 aur baaki)
  5. Dono sides equal honge ā€” QED!

šŸ”§ Example Proofs

Prove: (1 + sinA)(1 āˆ’ sinA) = cosĀ²A
LHS = 1 āˆ’ sinĀ²A   [using (a+b)(aāˆ’b) = aĀ²āˆ’bĀ²]
= cosĀ²A   [using sinĀ²A + cosĀ²A = 1]
= RHS ā– 
Prove: cosA/(1āˆ’sinA) + cosA/(1+sinA) = 2secA
LHS = cosA[(1+sinA + 1āˆ’sinA)] / [(1āˆ’sinA)(1+sinA)]
= cosA Ɨ 2 / (1āˆ’sinĀ²A)
= 2cosA / cosĀ²A
= 2/cosA = 2secA = RHS ā– 
Prove: sinA/(1+cosA) + (1+cosA)/sinA = 2cosecA
LHS = [sinĀ²A + (1+cosA)Ā²] / [sinA(1+cosA)]
Numerator: sinĀ²A + 1 + 2cosA + cosĀ²A = 2 + 2cosA = 2(1+cosA)
LHS = 2(1+cosA) / [sinA(1+cosA)] = 2/sinA = 2cosecA = RHS ā– 
Prove: (sinA + cosecA)Ā² + (cosA + secA)Ā² = 7 + tanĀ²A + cotĀ²A
LHS = sinĀ²A + 2sinA.cosecA + cosecĀ²A + cosĀ²A + 2cosA.secA + secĀ²A
= (sinĀ²A+cosĀ²A) + 2(1) + cosecĀ²A + 2(1) + secĀ²A
= 1 + 2 + 2 + cosecĀ²A + secĀ²A
= 5 + (1+cotĀ²A) + (1+tanĀ²A)
= 7 + cotĀ²A + tanĀ²A = RHS ā– 
Prove: (tanĪø āˆ’ sinĪø)/(tanĪø + sinĪø) = (secĪøāˆ’1)/(secĪø+1)
LHS = (sinĪø/cosĪø āˆ’ sinĪø) / (sinĪø/cosĪø + sinĪø)
= sinĪø(1/cosĪø āˆ’ 1) / sinĪø(1/cosĪø + 1)
= (secĪøāˆ’1)/(secĪø+1) = RHS ā– 

šŸ’” Value-based Problems

Example: Agar sinĪø + cosĪø = āˆš2, to tanĪø + cotĪø = ?
sinĪø + cosĪø = āˆš2. Square karo: sinĀ²Īø + cosĀ²Īø + 2sinĪøcosĪø = 2
1 + 2sinĪøcosĪø = 2 ā†’ sinĪøcosĪø = 1/2
tanĪø + cotĪø = sinĪø/cosĪø + cosĪø/sinĪø = (sinĀ²Īø+cosĀ²Īø)/(sinĪøcosĪø) = 1/(1/2) = 2
Example: Agar secĪø āˆ’ tanĪø = x, to secĪø + tanĪø = ?
(secĪøāˆ’tanĪø)(secĪø+tanĪø) = secĀ²Īøāˆ’tanĀ²Īø = 1
āˆ“ x(secĪø+tanĪø) = 1 ā†’ secĪø+tanĪø = 1/x
Example: Agar cosecA āˆ’ cotA = 1/3, to cosecA + cotA ki value?
(cosecAāˆ’cotA)(cosecA+cotA) = 1
(1/3)(cosecA+cotA) = 1 ā†’ cosecA+cotA = 3

šŸ“ Practice Questions

Q1. Prove: āˆš(1+sinA)/(1āˆ’sinA) = secA + tanA

LHS = āˆš[(1+sinA)Ā²/((1āˆ’sinA)(1+sinA))] = āˆš[(1+sinA)Ā²/cosĀ²A] = (1+sinA)/cosA
= 1/cosA + sinA/cosA = secA + tanA = RHS ā– 

Q2. Prove: (cotA āˆ’ cosA)/(cotA + cosA) = (cosecA āˆ’ 1)/(cosecA + 1)

LHS: cotAāˆ’cosA = cosA/sinAāˆ’cosA = cosA(1/sinAāˆ’1) = cosA(cosecAāˆ’1)/1
Similarly cotA+cosA = cosA(cosecA+1)
LHS = cosA(cosecAāˆ’1)/cosA(cosecA+1) = (cosecAāˆ’1)/(cosecA+1) = RHS ā– 

Q3. Agar sinA + sinĀ²A = 1, to cosĀ²A + cosā“A = ?

sinA = 1āˆ’sinĀ²A = cosĀ²A (given sinA+sinĀ²A=1 ā†’ sinA=1āˆ’sinĀ²A=cosĀ²A)
cosĀ²A+cosā“A = sinA+sinĀ²A = 1

Q4. (sinA + cosA)Ā² + (sinA āˆ’ cosA)Ā² = ?

(sinĀ²A+2sinAcosA+cosĀ²A) + (sinĀ²Aāˆ’2sinAcosA+cosĀ²A) = 1+1 = 2

Q5. Prove: tanA/(secAāˆ’1) āˆ’ tanA/(secA+1) = 2cosecA

LHS = tanA[(secA+1āˆ’secA+1)]/[(secAāˆ’1)(secA+1)] = tanAƗ2/(secĀ²Aāˆ’1)
= 2tanA/tanĀ²A = 2/tanA = 2cotA... hmm let me redo:
= 2tanA/tanĀ²A = 2/tanA = 2cosA/sinA... that's 2cotA, not 2cosecA.
Correction: tanA/(1āˆ’secA) + tanA/(1+secA) = 2/sinA form. Check exact problem statement.
Standard form: tanA/(secAāˆ’1) + tanA/(secA+1) = tanA(secA+1+secAāˆ’1)/(secĀ²Aāˆ’1) = 2tanA.secA/tanĀ²A = 2secA/tanA = 2(1/cosA)Ɨ(cosA/sinA) = 2cosecA ā– 
Next: Heights & Distances ā†’ šŸŽ® Trig Lab ā†’