2.4.6
Intuition Reciprocals kyun exist karte hainChhe trigonometric functions teen reciprocal pairs mein aate hain. Kyun? Kyunki ek right triangle mein, do sides ka ratio (jaise sine ke liye opposite/hypotenuse) ka ek natural "flip" hota hai — wahi sides ka ratio ulta order mein (hypotenuse/opposite). Yeh symmetry teen nayi functions banati hai: cosecant (cosec or csc) , secant (sec) , aur cotangent (cot) . Yeh "extra" nahi hain — yeh un expressions ko simplify karte hain jahan sin, cos, ya tan se divide karna baar baar aata hai.
Starting point: Right triangle with angle θ a b r
Step 1: Basic definitions
sin θ = r a , cos θ = r b , tan θ = b a
Yeh kyun? Yeh triangle ki shape ko angle θ
Step 2: Har ratio ko flip karo
s i n θ 1 = a / r 1 = a r = csc θ
Yeh step kyun? Algebraic reciprocal: p / q 1 = p q
Similarly:
c o s θ 1 = b r = sec θ t a n θ 1 = a b = cot θ
Step 3: Cotangent ka alternative form
Kyunki tan θ = c o s θ s i n θ cot θ = t a n θ 1 = s i n θ / c o s θ 1 = s i n θ c o s θ
Yeh kyun matter karta hai? cot θ = s i n θ c o s θ t a n θ 1
Worked example Example 1:csc 30° Given: sin 30° = 2 1 Solution:
csc 30° = s i n 30° 1 = 1/2 1 = 2
Yeh step kyun? Reciprocal definition ka direct application. Ek fraction se divide karna matlab uske reciprocal se multiply karna: 1/2 1 = 1 × 1 2 = 2
Verification: Ek 30-60-90 triangle mein jahan hypotenuse 2 aur opposite side 1 hai, opposite hypotenuse = 1 2 = 2
Worked example Example 2:c s c θ s e c θ Solution:
c s c θ s e c θ = 1/ s i n θ 1/ c o s θ
Yeh step kyun? Har function ko sin/cos ke terms mein uski definition se replace karo.
= c o s θ 1 × 1 s i n θ = c o s θ s i n θ
Yeh step kyun? Fraction se divide karna: b / c a = a × b c a = 1/ cos θ b = 1 c = sin θ
= tan θ
Answer: tan θ
Worked example Example 3: Agarcos θ = 5 3 θ sec θ csc θ Given: cos θ = 5 3 θ
Step 1: sec θ sec θ = c o s θ 1 = 3/5 1 = 3 5 Yeh step kyun? Direct reciprocal.
Step 2: Pythagorean identity use karke sin θ sin 2 θ + cos 2 θ = 1 sin 2 θ = 1 − ( 5 3 ) 2 = 1 − 25 9 = 25 16 sin θ = 5 4 θ
Yeh step kyun? csc θ sin θ
Step 3: csc θ csc θ = s i n θ 1 = 4/5 1 = 4 5 Answers: sec θ = 3 5 , csc θ = 4 5
Worked example Example 4: Prove karo1 + cot 2 θ = csc 2 θ Strategy: Pythagorean identity sin 2 θ + cos 2 θ = 1 sin 2 θ Step 1: Dono sides ko sin 2 θ s i n 2 θ s i n 2 θ + s i n 2 θ c o s 2 θ = s i n 2 θ 1 Yeh step kyun? Hum right side par csc 2 θ = 1/ sin 2 θ cot 2 θ = cos 2 θ / sin 2 θ
Step 2: Har term simplify karo
1 + s i n 2 θ c o s 2 θ = s i n 2 θ 1
Step 3: Reciprocal identities pehchano
1 + ( s i n θ c o s θ ) 2 = ( s i n θ 1 ) 2 1 + cot 2 θ = csc 2 θ
Yeh step kyun? Definition ke anusaar cot θ = cos θ / sin θ csc θ = 1/ sin θ
Conclusion: Identity proven. ✓
Common mistake Mistake 1:csc θ sin − 1 θ Galat soch: "csc θ sin θ csc θ = sin − 1 θ
Yeh sahi kyun lagta hai: Notation sin − 1 − 1
Fix yeh hai:
csc θ = s i n θ 1 multiplicative reciprocal hai (algebraic flip)sin − 1 θ arcsin θ inverse function hai (woh angle jiska sine θ Yeh bilkul alag hain: csc 30° = 2 sin − 1 ( 30° ) [ − 1 , 1 ]
Steel-man: − 1 f − 1 x − 1 arcsin
Common mistake Mistake 2: Yeh sochna ki reciprocals har jagah defined hainGalat calculation: "csc 0° = s i n 0° 1 = 0 1 = ∞
Yeh sahi kyun lagta hai: Value zero approach karne par "blow up" hoti hai, isliye infinity answer lagta hai.
Fix yeh hai: 0 1 undefined hai, infinity nahi. csc 0° θ = 0° Apna domain check karo: Koi reciprocal function compute karne se pehle verify karo ki base function non-zero hai.
Common mistake Mistake 3:cot Inefficient approach: c o s θ c o t θ s i n θ cot θ = t a n θ 1 tan θ = c o s θ s i n θ
Better approach: cot θ = s i n θ c o s θ c o s θ c o t θ s i n θ = c o s θ ( c o s θ / s i n θ ) ⋅ s i n θ = c o s θ c o s θ = 1
Yeh kyun matter karta hai: cot θ = s i n θ c o s θ
Mnemonic Reciprocal pairs mnemonic"Sine's Cousin, Cosine's Sister, Tangent's Twin"
Sine's Cousin = CoSec (csc) = s i n 1 Cosine's Sister = Sec (sec) = c o s 1 Tangent's Twin = Cot (cot) = t a n 1
Alternatively: "Reciprocals flip the ratio" — agar sin opp/hyp hai, toh csc hyp/opp hai.
Recall Feynman explanation (ek 12-saal ke bachche ko samjhao)Socho tum measure kar rahe ho ki ek ramp kitna steep hai. Tum bol sakte ho "har 3 meter aage ke liye, main 4 meter upar jaata hun" — yeh tangent jaisa hai (rise over run).
Lekin kabhi kabhi ise ulta bolna asaan hota hai: "har 4 meter upar jaane ke liye, main 3 meter aage jaata hun" — yeh cotangent hai! Same ramp, bas ulta describe kiya gaya.
Sine, cosine, aur tangent ke yeh "ulte" versions cosecant, secant, aur cotangent kehlaate hain. Yeh sirf flipped fractions hain. Inhe tab use karte hain jab sine/cosine/tangent se divide karna ek problem mein baar baar aaye — s i n θ 1 csc θ
Key idea: Agar sin 30° = 0.5 csc 30° = 2
#flashcards/maths
sin θ csc θ = s i n θ 1
cos θ sec θ = c o s θ 1
tan θ cot θ = t a n θ 1
cot θ sin θ cos θ cot θ = s i n θ c o s θ
csc θ Jab sin θ = 0 θ = 0° , 180° , 360° , … θ = nπ
sec θ Jab cos θ = 0 θ = 90° , 270° , … θ = 2 π + nπ
Agar sin θ = 5 4 csc θ csc θ = 4 5
csc θ sin θ csc θ sin θ = s i n θ 1 ⋅ sin θ = 1
csc θ sin − 1 θ csc θ = s i n θ 1 sin − 1 θ = arcsin θ
cot csc 1 + cot 2 θ = csc 2 θ
Ek right triangle mein, agar hypotenuse 13 aur opposite side 5 hai, toh csc θ csc θ = opposite hypotenuse = 5 13
c s c θ s e c θ c s c θ s e c θ = 1/ s i n θ 1/ c o s θ = c o s θ s i n θ = tan θ
Basic trig definitions — reciprocals teen primary functions ko extend karte hainPythagorean identities — 1 + cot 2 θ = csc 2 θ sin 2 + cos 2 = 1 sin 2 θ Complementary angles — csc ( 90° − θ ) = sec θ Trig equations — reciprocals tab appear hote hain jab csc x = 2 Calculus derivatives — d x d ( csc x ) = − csc x cot x
Yeh reciprocals master karo — yeh advanced trig identities aur calculus ki backbone hain.
Right triangle sides a b r
