3.1.10
Intuition Badi idea kya haiEk right triangle mein sirf teen fundamental ratios hote hain: sine, cosine, tangent. Baaki sab — cosecant, secant, cotangent — bas inhi mein se ek ko ulta kar do (reciprocal) ya do ko mil ke banao (quotient). Isliye tumhe 6 alag cheezein yaad nahi karni; 2–3 yaad karo aur baaki khud generate karo.
Definition Chheh trig ratios (right triangle se)Angle θ sin θ = H O , cos θ = H A , tan θ = A O reciprocals ke roop mein define hote hain:
csc θ = O H , sec θ = A H , cot θ = O A
Definition Reciprocal identitiescsc θ = == s i n θ == 1 , sec θ = == c o s θ == 1 , cot θ = == t a n θ == 1 sin θ csc θ = 1 cos θ sec θ = 1 tan θ cot θ = 1
Definition Quotient identitiestan θ = c o s θ == s i n θ == , cot θ = s i n θ == c o s θ ==
Hum sab kuch side-ratio definitions se derive karte hain — kuch bhi assume nahi kiya gaya.
Reciprocal — csc θ = 1/ sin θ
sin θ = H O ⟹ s i n θ 1 = O H . Ye step kyun? Fraction ko flip karna exactly "reciprocal" ka matlab hai. Aur H / O bilkul csc θ . Toh csc θ = 1/ sin θ sec = 1/ cos cot = 1/ tan
Quotient — tan θ = sin θ / cos θ
c o s θ s i n θ = A / H O / H . Ye step kyun? Definitions substitute karo. Numerator aur denominator dono mein H = H O ⋅ A H = A O = tan θ . Ye step kyun? H tan θ
Quotient — cot θ = cos θ / sin θ
s i n θ c o s θ = O / H A / H = O A = cot θ . cot θ = 1/ tan θ = 1/ ( sin / cos ) = cos / sin
use kaise karo (80/20 core)
Sirf sin cos
tan sin / cos csc , sec , cot sin , cos , tan
Worked example Example 1 — Ek point se saare chheh banaoθ ( 3 , 4 ) O = 4 A = 3 H = 3 2 + 4 2 = 5
sin θ = 4/5 cos θ = 3/5 Kyun? Definitions O / H A / H tan θ = cos θ sin θ = 3/5 4/5 = 3 4 Ye step kyun? Quotient identity — 5 csc θ = 1/ sin θ = 5/4 sec θ = 5/3 cot θ = 3/4 Kyun? Reciprocals; bas flip karo.
Worked example Example 3 — Expression simplify karosec θ tan θ s e c θ t a n θ = 1/ c o s θ s i n θ / c o s θ = c o s θ s i n θ ⋅ cos θ = sin θ . Har step kyun? sin , cos cos sin / cos
Recall Forecast-then-VerifyPadhne se pehle: predict karo csc θ cot θ 1/ sin cos / sin = sin cos ⋅ sin = cos θ . cos θ
cot θ = 1/ tan θ cot θ = cos θ / sin θ sin / cos Error sahi kyun lagti hai: cot cos Fix: cot = 1/ tan = 1/ ( sin / cos ) = cos / sin cos cot A / O tan sin cot cos
sec θ = 1/ sin θ Sahi kyun lagta hai: "sec" aur "sin" dono 's' se shuru hote hain, toh brain inhe pair kar leta hai.
Fix: Pairing cross hoti hai: sec ↔ cos csc ↔ sin cosecant pairs with sine ('co' of cosecant mein ek 's' chhupi hai... bas neeche wala mnemonic yaad karo).
csc θ = sin − 1 θ Sahi kyun lagta hai: x − 1 sin − 1 Fix: sin − 1 inverse function hai (arcsin), reciprocal NAHI. Reciprocal hai ( sin θ ) − 1 = csc θ
Mnemonic Pairing & quotient memory hook"Jo 'C' se shuru hota hai wo us se pair karta hai jo match NAHI karta (co-swap):"
cosec ant ↔ s inesec ant ↔ cos inecot angent ↔ tan gent (ye pair normally match karta hai)
Quotient: "TAN = SIN over COS" (alphabetically S pehle, C baad mein = numerator). cot
Recall Feynman: ek 12-saal ke bacche ko samjhaoEk triangle socho. Ek kone ki "steepness" teen taraon se measure kar sakte ho, do sides ko ek saath use karke — inhe sine, cosine, tangent kaho. Baaki teen fancy names (cosecant, secant, cotangent) naye nahi hain — ye wohi numbers hain ulte karke , jaise 3 2 2 3
csc θ sin θ csc θ = 1/ sin θ
sec θ cos θ sec θ = 1/ cos θ
cot θ tan θ cot θ = 1/ tan θ
tan θ tan θ = sin θ / cos θ
cot θ cot θ = cos θ / sin θ
sec cos
csc sin
tan θ / sec θ sin θ
cot θ ⋅ sin θ cos θ cot = cos / sin
Kya csc θ sin − 1 θ Nahi — sin − 1 csc ( sin θ ) − 1
Side ratios se tan θ = sin / cos A / H O / H = A O = tan θ
Agar sin θ = 2/3 csc θ 3/2
Pythagorean identities — jab sirf sin cos Unit circle definitions — in ratios ko right triangles se aage saare angles tak extend karta hai.Simplifying trigonometric expressions — "sin & cos mein convert karo" strategy yahan rehti hai.Right triangle trigonometry (SOH-CAH-TOA) — saaro chheh definitions ka source.Inverse trigonometric functions — sin − 1 csc
csc=1/sin sec=1/cos cot=1/tan
