There are only three fundamental ratios in a right triangle: sine, cosine, tangent. Everything else — cosecant, secant, cotangent — is just one of these turned upside down (reciprocal) or built from two of them (quotient). So you never memorise 6 independent things; you memorise 2–3 and generate the rest.
Definition The six trig ratios (from a right triangle)For an angle θ \theta θ sin θ = O H , cos θ = A H , tan θ = O A \sin\theta = \frac{O}{H}, \qquad \cos\theta = \frac{A}{H}, \qquad \tan\theta = \frac{O}{A} sin θ = H O , cos θ = H A , tan θ = A O reciprocals :
csc θ = H O , sec θ = H A , cot θ = A O \csc\theta = \frac{H}{O}, \qquad \sec\theta = \frac{H}{A}, \qquad \cot\theta = \frac{A}{O} csc θ = O H , sec θ = A H , cot θ = O A
Definition Reciprocal identitiescsc θ = 1 = = sin θ = = , sec θ = 1 = = cos θ = = , cot θ = 1 = = tan θ = = \csc\theta = \frac{1}{==\sin\theta==}, \qquad \sec\theta = \frac{1}{==\cos\theta==}, \qquad \cot\theta = \frac{1}{==\tan\theta==} csc 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 \sin\theta\,\csc\theta = 1 sin θ csc θ = 1 cos θ sec θ = 1 \cos\theta\,\sec\theta = 1 cos θ sec θ = 1 tan θ cot θ = 1 \tan\theta\,\cot\theta = 1 tan θ cot θ = 1
Definition Quotient identitiestan θ = = = sin θ = = cos θ , cot θ = = = cos θ = = sin θ \tan\theta = \frac{==\sin\theta==}{\cos\theta}, \qquad \cot\theta = \frac{==\cos\theta==}{\sin\theta} tan θ = c o s θ == s i n θ == , cot θ = s i n θ == c o s θ ==
We derive everything from the side-ratio definitions — nothing is assumed.
Reciprocal — derive csc θ = 1 / sin θ \csc\theta = 1/\sin\theta csc θ = 1/ sin θ
sin θ = O H ⟹ 1 sin θ = H O . \sin\theta = \frac{O}{H} \;\Longrightarrow\; \frac{1}{\sin\theta} = \frac{H}{O}. sin θ = H O ⟹ s i n θ 1 = O H . Why this step? Flipping a fraction is exactly what "reciprocal" means. And H / O H/O H / O precisely the definition of csc θ \csc\theta csc θ . So csc θ = 1 / sin θ \csc\theta = 1/\sin\theta csc θ = 1/ sin θ sec = 1 / cos \sec = 1/\cos sec = 1/ cos cot = 1 / tan \cot = 1/\tan cot = 1/ tan
Quotient — derive tan θ = sin θ / cos θ \tan\theta = \sin\theta/\cos\theta tan θ = sin θ / cos θ
sin θ cos θ = O / H A / H . \frac{\sin\theta}{\cos\theta} = \frac{O/H}{A/H}. c o s θ s i n θ = A / H O / H . Why this step? Substitute the definitions. Both numerator and denominator carry an H H H = O H ⋅ H A = O A = tan θ . = \frac{O}{H}\cdot\frac{H}{A} = \frac{O}{A} = \tan\theta. = H O ⋅ A H = A O = tan θ . Why this step? The H H H tan θ \tan\theta tan θ
Quotient — derive cot θ = cos θ / sin θ \cot\theta = \cos\theta/\sin\theta cot θ = cos θ / sin θ
cos θ sin θ = A / H O / H = A O = cot θ . \frac{\cos\theta}{\sin\theta} = \frac{A/H}{O/H} = \frac{A}{O} = \cot\theta. 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 \cot\theta = 1/\tan\theta = 1/(\sin/\cos) = \cos/\sin cot θ = 1/ tan θ = 1/ ( sin / cos ) = cos / sin
use them (the 80/20 core)
Memorise only sin \sin sin cos \cos cos
Need tan \tan tan sin / cos \sin/\cos sin / cos
Need csc , sec , cot \csc,\sec,\cot csc , sec , cot sin , cos , tan \sin,\cos,\tan sin , cos , tan
Worked example Example 1 — Build all six from a pointA point on the terminal side of θ \theta θ ( 3 , 4 ) (3,4) ( 3 , 4 ) O = 4 O=4 O = 4 A = 3 A=3 A = 3 H = 3 2 + 4 2 = 5 H=\sqrt{3^2+4^2}=5 H = 3 2 + 4 2 = 5
sin θ = 4 / 5 \sin\theta = 4/5 sin θ = 4/5 cos θ = 3 / 5 \cos\theta = 3/5 cos θ = 3/5 Why? Definitions O / H O/H O / H A / H A/H A / H tan θ = sin θ cos θ = 4 / 5 3 / 5 = 4 3 \tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{4/5}{3/5} = \dfrac{4}{3} tan θ = cos θ sin θ = 3/5 4/5 = 3 4 Why this step? Quotient identity — the 5 5 5 csc θ = 1 / sin θ = 5 / 4 \csc\theta = 1/\sin\theta = 5/4 csc θ = 1/ sin θ = 5/4 sec θ = 5 / 3 \sec\theta = 5/3 sec θ = 5/3 cot θ = 3 / 4 \cot\theta = 3/4 cot θ = 3/4 Why? Reciprocals; just flip.
Worked example Example 2 — Given one ratio, find anotherGiven sin θ = 2 3 \sin\theta = \tfrac{2}{3} sin θ = 3 2 csc θ \csc\theta csc θ cot θ \cot\theta cot θ
csc θ = 1 / sin θ = 3 / 2 \csc\theta = 1/\sin\theta = 3/2 csc θ = 1/ sin θ = 3/2 Why? Reciprocal — instant, no triangle needed.For cot \cot cot cos \cos cos cos θ = 1 − sin 2 θ = 1 − 4 / 9 = 5 / 3 \cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-4/9}=\sqrt{5}/3 cos θ = 1 − sin 2 θ = 1 − 4/9 = 5 /3 Why this step? Pythagorean identity supplies the missing ratio.
cot θ = cos θ sin θ = 5 / 3 2 / 3 = 5 2 \cot\theta = \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sqrt5/3}{2/3} = \dfrac{\sqrt5}{2} cot θ = sin θ cos θ = 2/3 5 /3 = 2 5 Why? Quotient identity; the 3 3 3
Worked example Example 3 — Simplify an expressionSimplify tan θ sec θ \dfrac{\tan\theta}{\sec\theta} sec θ tan θ tan θ sec θ = sin θ / cos θ 1 / cos θ = sin θ cos θ ⋅ cos θ = sin θ . \frac{\tan\theta}{\sec\theta} = \frac{\sin\theta/\cos\theta}{1/\cos\theta} = \frac{\sin\theta}{\cos\theta}\cdot\cos\theta = \sin\theta. 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 . Why each step? Rewrite in sin , cos \sin,\cos sin , cos cos \cos cos sin / cos \sin/\cos sin / cos
Recall Forecast-then-VerifyBefore reading: predict cot θ csc θ \dfrac{\cot\theta}{\csc\theta} csc θ cot θ cos / sin 1 / sin = cos sin ⋅ sin = cos θ . \dfrac{\cos/\sin}{1/\sin} = \dfrac{\cos}{\sin}\cdot\sin = \cos\theta. 1/ sin cos / sin = sin cos ⋅ sin = cos θ . cos θ \cos\theta cos θ
cot θ = 1 / tan θ \cot\theta = 1/\tan\theta cot θ = 1/ tan θ cot θ = cos θ / sin θ \cot\theta = \cos\theta/\sin\theta cot θ = cos θ / sin θ sin / cos \sin/\cos sin / cos Why the error feels right: cot \cot cot looks like it starts with "co-" like cos \cos cos Fix: cot = 1 / tan = 1 / ( sin / cos ) = cos / sin \cot = 1/\tan = 1/(\sin/\cos) = \cos/\sin cot = 1/ tan = 1/ ( sin / cos ) = cos / sin cos \cos cos cot \cot cot A / O A/O A / O tan \tan tan sin \sin sin cot \cot cot cos \cos cos
sec θ = 1 / sin θ \sec\theta = 1/\sin\theta sec θ = 1/ sin θ Why it feels right: "sec" and "sin" both start with 's', so the brain pairs them.
Fix: The pairing is cross : sec ↔ cos \sec\leftrightarrow\cos sec ↔ cos csc ↔ sin \csc\leftrightarrow\sin csc ↔ sin doesn't match is the trap. Use: cosecant pairs with sine (the 'co' of cosecant hides an 's'... just memorise the mnemonic below).
csc θ = sin − 1 θ \csc\theta = \sin^{-1}\theta csc θ = sin − 1 θ Why it feels right: x − 1 x^{-1} x − 1 sin − 1 \sin^{-1} sin − 1 looks like it should too.
Fix: sin − 1 \sin^{-1} sin − 1 inverse function (arcsin), NOT the reciprocal. Reciprocal is ( sin θ ) − 1 = csc θ (\sin\theta)^{-1} = \csc\theta ( sin θ ) − 1 = csc θ
Mnemonic Pairing & quotient memory hook"The one that starts with C pairs with the one that DOESN'T (co-swap):"
cosec ant ↔ s inesec ant ↔ cos inecot angent ↔ tan gent (this pair matches normally)
Quotient: "TAN = SIN over COS" (alphabetical S before C = numerator). cot \cot cot
Recall Feynman: explain to a 12-year-oldImagine a triangle. You can measure a corner's "steepness" three ways using two sides at a time — call them sine, cosine, tangent. The other three fancy names (cosecant, secant, cotangent) aren't new — they're just the same numbers flipped over , like turning 2 3 \tfrac{2}{3} 3 2 3 2 \tfrac{3}{2} 2 3
What is csc θ \csc\theta csc θ sin θ \sin\theta sin θ csc θ = 1 / sin θ \csc\theta = 1/\sin\theta csc θ = 1/ sin θ What is sec θ \sec\theta sec θ cos θ \cos\theta cos θ sec θ = 1 / cos θ \sec\theta = 1/\cos\theta sec θ = 1/ cos θ What is cot θ \cot\theta cot θ tan θ \tan\theta tan θ cot θ = 1 / tan θ \cot\theta = 1/\tan\theta cot θ = 1/ tan θ State the quotient identity for tan θ \tan\theta tan θ tan θ = sin θ / cos θ \tan\theta = \sin\theta/\cos\theta tan θ = sin θ / cos θ State the quotient identity for cot θ \cot\theta cot θ cot θ = cos θ / sin θ \cot\theta = \cos\theta/\sin\theta cot θ = cos θ / sin θ Which ratio does sec \sec sec cos \cos cos (secant = 1/cosine)
Which ratio does csc \csc csc sin \sin sin (cosecant = 1/sine)
Simplify tan θ / sec θ \tan\theta/\sec\theta tan θ / sec θ Simplify cot θ ⋅ sin θ \cot\theta \cdot \sin\theta cot θ ⋅ sin θ cos θ \cos\theta cos θ (since
cot = cos / sin \cot=\cos/\sin cot = cos / sin )
Is csc θ \csc\theta csc θ sin − 1 θ \sin^{-1}\theta sin − 1 θ No —
sin − 1 \sin^{-1} sin − 1 is arcsin (inverse function);
csc \csc csc is the reciprocal
( sin θ ) − 1 (\sin\theta)^{-1} ( sin θ ) − 1 .
Derive tan θ = sin / cos \tan\theta=\sin/\cos tan θ = sin / cos O / H A / H = O A = tan θ \frac{O/H}{A/H}=\frac{O}{A}=\tan\theta A / H O / H = A O = tan θ (H cancels).
If sin θ = 2 / 3 \sin\theta=2/3 sin θ = 2/3 csc θ \csc\theta csc θ
Pythagorean identities — supplies the missing cos \cos cos sin \sin sin Unit circle definitions — extends these ratios beyond right triangles to all angles.Simplifying trigonometric expressions — "convert to sin & cos" strategy lives here.Right triangle trigonometry (SOH-CAH-TOA) — the source of all six definitions.Inverse trigonometric functions — contrast sin − 1 \sin^{-1} sin − 1 csc \csc csc
csc=1/sin sec=1/cos cot=1/tan
Intuition Hinglish mein samjhoDekho, trigonometry mein sirf teen basic ratios hote hain: sine, cosine aur tangent. Baaki jo teen naam hain — cosecant, secant, cotangent — woh koi naye monster nahi hain. Woh bas inhi ka ulta (reciprocal) hain. Jaise csc θ = 1 / sin θ \csc\theta = 1/\sin\theta csc θ = 1/ sin θ sec θ = 1 / cos θ \sec\theta = 1/\cos\theta sec θ = 1/ cos θ cot θ = 1 / tan θ \cot\theta = 1/\tan\theta cot θ = 1/ tan θ
Quotient identity ka matlab hai ki tangent actually chhupa hua sin / cos \sin/\cos sin / cos cos / sin \cos/\sin cos / sin TAN mein SIN upar hota hai , aur cot ulta. Proof simple hai — tan = ( O / H ) ÷ ( A / H ) \tan = (O/H)\div(A/H) tan = ( O / H ) ÷ ( A / H ) O / A O/A O / A
Do bade traps hain. Pehla: log sec = 1 / sin \sec = 1/\sin sec = 1/ sin cross hoti hai: cosecant ka sine, secant ka cosine. Doosra trap: csc θ \csc\theta csc θ sin − 1 θ \sin^{-1}\theta sin − 1 θ sin − 1 \sin^{-1} sin − 1
Jab bhi koi complicated trig expression simplify karna ho, ek universal trick yaad rakho: sab kuch sin aur cos mein convert kar do , phir jo cancel ho raha hai use kaat do. Jaise tan / sec = ( sin / cos ) / ( 1 / cos ) = sin \tan/\sec = (\sin/\cos)/(1/\cos) = \sin tan / sec = ( sin / cos ) / ( 1/ cos ) = sin
