Introduction to Trigonometry 📐

Right triangle ke ratios, standard angles ki table, aur reciprocal relations.

🌟 Trigonometry Kya Hai?

🏗️ Engineer kaise karte hain?

Egypt ke pyramids banane waale, Eiffel Tower design karne waale, aur GPS satellites track karne waale — sab trigonometry use karte hain. Ek angle aur ek side se, baaki sab nikal sakte hain. "Trigonometry" = Greek words "trigon" (triangle) + "metron" (measure) = triangle measurement!

Trigonometry right-angled triangle ke angles aur sides ke relationships ko study karta hai.
Sirf ek angle aur ek side know ho to — baaki sab nikalte hain!

📐 Right Triangle — Parts Yaad Karo

θ A B C Adjacent (BC) Opposite (AC) Hypotenuse (AB) Right angle at C θ is angle at B
For angle θ at vertex B:
Opposite side = side facing angle θ = AC
Adjacent side = side next to angle θ (not hypotenuse) = BC
Hypotenuse = side opposite to 90° = AB (longest side)

🎵 SOH-CAH-TOA — Yaad Karne Ka Formula

SOH CAH TOA

Sin = Opposite / Hypotenuse  |  Cos = Adjacent / Hypotenuse  |  Tan = Opposite / Adjacent

📊 6 Trigonometric Ratios

sin θ
= Opp/Hyp = P/H
Reciprocal: cosec θ = H/P
cos θ
= Adj/Hyp = B/H
Reciprocal: sec θ = H/B
tan θ
= Opp/Adj = P/B
Reciprocal: cot θ = B/P
cosec θ
= Hyp/Opp = H/P
= 1/sin θ
sec θ
= Hyp/Adj = H/B
= 1/cos θ
cot θ
= Adj/Opp = B/P
= 1/tan θ = cos/sin
tan θ = sin θ / cos θ    cot θ = cos θ / sin θ
cosec × sin = 1    sec × cos = 1    cot × tan = 1
Example 1: Right △ABC mein ∠C=90°, BC=3, AB=5. Angle A ke liye sab ratios.
AC = √(AB²−BC²) = √(25−9) = 4
For ∠A: Opp=BC=3, Adj=AC=4, Hyp=AB=5
sin A = 3/5, cos A = 4/5, tan A = 3/4
cosec A = 5/3, sec A = 5/4, cot A = 4/3
Example 2: sin A = 1/2. Baaki ratios nikalo (A acute angle).
sin A = P/H = 1/2. Let P=1, H=2. B = √(4−1) = √3
cos A = √3/2, tan A = 1/√3 = √3/3
cosec A = 2, sec A = 2/√3 = 2√3/3, cot A = √3

📐 Standard Angles Table

Memory Trick for sin: sin 0° = √0/2 = 0, sin 30° = √1/2 = 1/2, sin 45° = √2/2, sin 60° = √3/2, sin 90° = √4/2 = 1
cos: Same sequence reverse mein → cos 0°=1, cos 30°=√3/2, cos 45°=√2/2, cos 60°=1/2, cos 90°=0
Angle →0°30°45°60°90°
sin01/21/√2√3/21
cos1√3/21/√21/20
tan01/√31√3Not defined
cosecNot def2√22/√31
sec12/√3√22Not def
cotNot def√311/√30
Example: Evaluate: 2tan²45° + cos²30° − sin²60°
= 2(1)² + (√3/2)² − (√3/2)²
= 2(1) + 3/4 − 3/4
= 2
Example: sin60°cos30° + cos60°sin30° = ?
= (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1
(Ye sin(60°+30°) = sin90° = 1 hai — addition formula!)

🔄 Complementary Angles

Complementary angle relations (A+B=90°)
sin A = cos(90°−A) = cos B
cos A = sin(90°−A) = sin B
tan A = cot(90°−A) = cot B
cosec A = sec(90°−A)
sec A = cosec(90°−A)
cot A = tan(90°−A)
Example: Evaluate: sin35°/cos55° + cos55°/sin35°
cos55° = sin35°, sin35° = cos55°
= sin35°/sin35° + sin35°/sin35° = 1 + 1 = 2
Example: tan5° × tan25° × tan45° × tan65° × tan85°
tan5°=cot85°, tan25°=cot65°, tan45°=1
= (tan5°×cot5°) × (tan25°×cot25°) × 1 = 1×1×1 = 1

📏 Finding Trigonometric Ratios

Example: ∠B=90° triangle mein, tan A = 5/12. sin A aur cos A nikalo.
tan A = Opp/Adj = BC/AB = 5/12. Let BC=5k, AB=12k.
Hypotenuse AC = √(25k²+144k²) = 13k
sin A = BC/AC = 5/13, cos A = AB/AC = 12/13
Example: 15 cot A = 8. sin A aur sec A nikalo.
cot A = 8/15 → B/P = 8/15. Let B=8, P=15. H=√(64+225)=√289=17
sin A = P/H = 15/17, sec A = H/B = 17/8

📝 Practice Questions

Q1. △PQR mein ∠Q=90°, PQ=6, PR=10. sin R, cos R, tan P nikalo.

QR = √(100−36) = 8. sin R = PQ/PR = 6/10 = 3/5. cos R = QR/PR = 4/5. tan P = QR/PQ = 8/6 = 4/3

Q2. Evaluate: sin²60° + cos²30° − tan²45°

(√3/2)² + (√3/2)² − 1² = 3/4 + 3/4 − 1 = 3/2 − 1 = 1/2

Q3. cos A = 3/5. Baki sab ratios nikalo.

cos A = B/H = 3/5. P = √(25−9) = 4. sin A=4/5, tan A=4/3, cosec A=5/4, sec A=5/3, cot A=3/4

Q4. Simplify: (sin47°/cos43°)² + (cos43°/sin47°)² − 4cos²45°

cos43°=sin47° (complementary). So = (sin47°/sin47°)² + (sin47°/sin47°)² − 4(1/2) = 1+1−2 = 0

Q5. cos9A = sin A (A acute). A nikalo.

sin A = sin(90°−9A) → A = 90°−9A → 10A = 90° → A = 9°
Next: Trig Identities → 🎮 Trig Lab →