Inverse Trigonometric Functions: Class 12 Maths Chapter 2 Hinglish mein

Mathematics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

65 min
48 practice questions
16 MCQs

Is chapter mein aap kya seekhenge

Samjhao ki har trigonometric function ko inverse milne se pehle restrict kyun karna padta hai, aur uski principal value branch ke liye use hone wala interval batao.
Chhahon inverse trigonometric functions ka domain aur principal range batao, aur ek branch ko line y = x mein reflect karke uska graph sketch karo.
Standard arguments par, negative arguments bhi, inverse trigonometric functions ki principal values nikaalo.
sin(sin⁻¹x) = x aur sin⁻¹(sin x) = x, aur unke jaise doosre results ko sirf unhi intervals par use karo jahan woh sahi hote hain, aur kisi bhi angle a ke liye sin⁻¹(sin a) jaise expressions evaluate karo.
Sahi substitution se inverse trigonometric expressions simplify karo, aur naye angle ka interval dhyaan mein rakho.
Right triangles aur addition formulae se inverse trigonometric functions ke sums aur compositions evaluate aur prove karo, aur unwale simple equations solve karo.

NCERT Mathematics Class 12, Chapter 2, pp. 18–33 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

2.1 Introduction

Phone ko tilt karo to uski screen tilt ko degrees mein dikha sakti hai. Andar laga accelerometer woh angle nahi naapta. Woh yeh naapta hai ki gravity ka kitna fraction phone ke lambe edge ke along lagta hai, aur woh fraction tilt ka sine hai. Degrees dikhane ke liye software ko sine ulta chalana padta hai: $\sin θ = 0.5$ se use $θ$ wapas nikaalna hota hai.

Kisi function ko ulta chalana uske inverse ka kaam hai, aur Chapter 1 ne humein bataya tha ki yeh kaam kab ho sakta hai: function one-one aur onto hona chahiye. Sine pehle test mein buri tarah fail hota hai: $θ = \frac{π}{6}$ , $\frac{5 π}{6}$ , $\frac{13 π}{6}$ aur infinitely many doosre angles ke liye $\sin θ = 0.5$ hai (Fig. 2.1). Yeh ek many-one function hai, aur baaki paanch trigonometric functions bhi aise hi hain.

4/4 $- \frac{π}{2} \leq x \leq \frac{π}{2}$ par bilkul ek hi answer hai, $x = \frac{π}{6}$ .

Fig. 2.1Jab tak domain chhota nahi kiya jaata, sine ka ek output kai inputs se aata hai.

Step 1 mein y = sin x ka graph lagbhag minus 2 pi se 2 pi tak dikhta hai, minus 1 aur 1 ke beech ek wave. Step 2 mein height one half par ek dashed horizontal line aati hai. Step 3 un chaar jagahon ko mark karta hai jahan line wave se milti hai, saath mein likha hai kai angles. Step 4 mein wave fade ho jaati hai, bas minus pi by 2 se pi by 2 tak ka piece bachta hai; woh strip shade hoti hai aur x = pi by 6 par sirf ek crossing rehti hai, jis par likha hai ek angle.

Ilaaj yeh hai ki domain ka kuch hissa chhod do. Agar hum tay kar lein ki sirf $- \frac{π}{2}$ se $\frac{π}{2}$ tak ke angles dekhenge, to unmein se bilkul ek ka sine $0.5$ hai, aur ulte sawaal ka ek hi answer milta hai. Yeh chapter chhahon trigonometric functions ke liye aisa choice karta hai, usse bane inverse functions ke graphs banata hai aur unke rules nikaalta hai.

Yeh inverses sirf kisi technical problem ka jugaad nahi hain. Calculus mein yeh $\int \frac{d x}{1 + x^{2}}$ jaise integrals ke answers ke roop mein aate hain, aur surveyors, physicists aur engineers inhe tab use karte hain jab naapi gayi lengths se koi angle wapas nikaalna ho: road ka gradient, dish antenna ki elevation, robot arm ke joint angles.

Quick check

2 mein se 1

Saare real numbers par liya gaya

\sin x

ka inverse kyun nahi hota?

Iska graph ek curve hai, straight line nahiKuch angles ke liye yeh negative values leta haiAlag-alag angles same sine dete hain, isliye yeh one-one nahi haiYeh

x = \frac{π}{2}

par undefined hai

Answer dekho

Answer (C) Alag-alag angles same sine dete hain, isliye yeh one-one nahi hai

Inverse ko har output ko wapas ek input par bhejna hota hai, lekin

\sin \frac{π}{6} = \sin \frac{5 π}{6}

. Curved graph aur negative values koi rukaawat nahi hain (

x^{3}

ke baare mein socho), aur sine

\frac{π}{2}

par defined hai; wahan undefined tangent hota hai.

2.2 Basic Concepts

Is section mein saara kaam pichhli classes ke do facts karte hain: har trigonometric function kahan rehta hai, aur kisi function ko invert karne ke liye kya chahiye. Pehla fact Table 2.1 mein ikattha kiya gaya hai.

Table 2.1Chhahon trigonometric functions apne natural domains par

Function	Domain	Range
$\sin x$	$𝐑$	$[- 1, 1]$
$\cos x$	$𝐑$	$[- 1, 1]$
$\tan x$	$𝐑 - {(2 n + 1) \frac{π}{2} : n \in 𝐙}$	$𝐑$
$\cot x$	$𝐑 - {n π : n \in 𝐙}$	$𝐑$
$\sec x$	$𝐑 - {(2 n + 1) \frac{π}{2} : n \in 𝐙}$	$𝐑 - (- 1, 1)$
$cosec x$	$𝐑 - {n π : n \in 𝐙}$	$𝐑 - (- 1, 1)$

Chhahon functions period $2 π$ ya $π$ ke saath khud ko repeat karte hain, isliye koi bhi apne natural domain par one-one nahi hai. Ab doosra fact.

Definition

Inverse function

Woh function $f^{- 1}$ jo kisi one-one aur onto function $f$ ka kaam ulta kar deta hai: yeh har output $f (x)$ ko wapas input $x$ par bhejta hai. Inputs aur outputs apni jagah badal lete hain, isliye $f^{- 1}$ ka domain $f$ ki range hai, aur range domain.

Example $𝐑$ par $f (x) = 2 x + 1$ ka inverse $f^{- 1} (y) = \frac{y - 1}{2}$ hai.

Undo karo, phir redo

\begin{array}{c} f^{- 1} (f (x)) = x \\ f (f^{- 1} (y)) = y \end{array}

Yeh $X$ ke har $x$ aur $Y$ ke har $y$ ke liye sahi hain. Undo karna dono directions mein kaam karta hai, isliye $f$ bhi $f^{- 1}$ ka inverse hai. Isse har trigonometric function ka plan tay ho jaata hai: aisa interval chuno jis par woh one-one ho aur phir bhi apni range ki har value le, phir us piece ko invert karo.

Inverse sine

$[- \frac{π}{2}, \frac{π}{2}]$ par sine function $- 1$ se $1$ tak lagaataar badhta hai aur beech ki har value se bilkul ek baar guzarta hai. Is interval par restrict karne se $\sin : [- \frac{π}{2}, \frac{π}{2}] \to [- 1, 1]$ one-one aur onto hai, isliye iska inverse hai. Aise chhote kiye gaye interval ko restricted domain kehte hain.

Sirf yahi interval kaam nahi karta. $[\frac{π}{2}, \frac{3 π}{2}]$ par sine $1$ se $- 1$ tak lagaataar girta hai aur $[- \frac{5 π}{2}, - \frac{3 π}{2}]$ par phir badhta hai; asal mein $k$ integer ho to koi bhi interval $[(2 k - 1) \frac{π}{2}, (2 k + 1) \frac{π}{2}]$ chalega. $[0, π]$ jaisa interval kaam nahi karta, kyunki uspar curve upar jaakar wapas neeche aa jaata hai (Fig. 2.2).

2/4 $[- \frac{π}{2}, \frac{π}{2}]$ par sine $- 1$ se $1$ tak badhta hai: one-one aur $[- 1, 1]$ par onto.

Fig. 2.2Sine curve ke kaun se pieces invert kiye ja sakte hain?NCERT Fig. 2.1 · p. 20

Step 1 mein minus 2 pi se 2 pi tak sine curve dikhta hai. Step 2 minus pi by 2 se pi by 2 tak ki strip shade karta hai aur us piece ko mota aur blue banata hai, jo minus 1 se 1 tak badhta hai, us par likha hai principal piece. Step 3 iske badle pi by 2 se 3 pi by 2 tak ka piece orange mein highlight karta hai, jo 1 se minus 1 tak girta hai, us par likha hai yeh bhi one-one. Step 4 0 se pi tak ka piece dotted red line se highlight karta hai, saath mein height one half par ek horizontal line hai jo use do baar milti hai, us par likha hai do inputs: fail.

Har chalne wala interval ek alag inverse deta hai, jise $\sin^{- 1}$ ki ek branch kehte hain (padho “sine inverse” ya “arc sine”). Har branch ka domain $[- 1, 1]$ hai; branches ki range alag hoti hai. Agreement se hum woh branch use karte hain jiski range $[- \frac{π}{2}, \frac{π}{2}]$ hai. Yeh $\sin^{- 1}$ ki principal value branch hai.

Definition

Principal value branch

Jin intervals par koi trigonometric function one-one aur onto hai, unmein se woh interval jo convention se uske inverse ki range ke roop mein fix kiya gaya hai. $\sin^{- 1}$ ke liye yeh $[- \frac{π}{2}, \frac{π}{2}]$ hai.

ExampleRange $[\frac{π}{2}, \frac{3 π}{2}]$ wali $\sin^{- 1}$ ki branch ek valid branch hai, lekin principal nahi.

Inverse sine

\begin{array}{c} \sin^{- 1} : [- 1, 1] \\ \to [- \frac{π}{2}, \frac{π}{2}] \end{array}

Shabdon mein, $y = \sin^{- 1} x$ ka matlab hai ki $\sin y = x$ aur $- \frac{π}{2} \leq y \leq \frac{π}{2}$ . Yeh hamara pehla inverse trigonometric function hai, aur $y$ par lagi condition iski definition ka hissa hai. Jaise $\sin \frac{5 π}{6} = \frac{1}{2}$ , phir bhi $\sin^{- 1} \frac{1}{2} = \frac{π}{6}$ hai, $\frac{5 π}{6}$ nahi, kyunki sirf $\frac{π}{6}$ principal branch mein aata hai. Isse turant do facts milte hain. Pehla, har $x \in [- 1, 1]$ ke liye $\sin (\sin^{- 1} x) = x$ . Doosra, har $x \in [- \frac{π}{2}, \frac{π}{2}]$ ke liye $\sin^{- 1} (\sin x) = x$ , aur sirf wahin.

Graph banana. Maan lo $f$ invertible hai aur $f (u) = v$ . Tab $f^{- 1} (v) = u$ , isliye jab bhi $(u, v)$ $f$ ke graph par ho, point $(v, u)$ $f^{- 1}$ ke graph par hota hai. Har point ke coordinates swap karna plane ko line $y = x$ mein reflect karne jaisa hi hai. Isliye $y = \sin^{- 1} x$ ka graph, $y = \sin x$ ke principal piece ki us line mein mirror image hai (Fig. 2.3).

3/3Har point isi tarah reflect hota hai. Image $y = \sin^{- 1} x$ hai, $(- 1, - \frac{π}{2})$ se $(1, \frac{π}{2})$ tak.

Fig. 2.3

y = x

mein reflect karne se har point ke coordinates swap ho jaate hain.NCERT Fig. 2.1 (iii) · p. 20

Dono axes par equal scales, lagbhag minus 2 se 2 tak. Step 1 mein y = sin x ka blue piece minus pi by 2 se pi by 2 tak dikhta hai, end point P (pi by 2, 1) par. Step 2 dashed mirror line y = x aur image point P prime (1, pi by 2) par jodta hai, jo P se mirror ke across ek dotted segment se juda hai. Step 3 poori mirror image ko dashed red curve ke roop mein jodta hai, y = sin inverse x, jo (minus 1, minus pi by 2) se (1, pi by 2) tak jaata hai aur dono ends par steep hai.

Naye graph ko kisi bhi aur graph ki tarah padho. Yeh sirf $- 1 \leq x \leq 1$ ke liye exist karta hai, $- \frac{π}{2}$ se $\frac{π}{2}$ tak badhta hai, aur apne ends par sabse steep hai, theek wahin jahan sine curve sabse flat tha. $\sin^{- 1}$ ki koi doosri branch kisi doosre piece ki mirror image hoti, jo $y$ -axis par upar ya neeche hoti.

Inverse cosine

Cosine ek even function hai, isliye yeh $0$ ke around symmetric kisi bhi interval par one-one nahi ho sakta. Iske badle hum $[0, π]$ use karte hain, jis par cosine $1$ se $- 1$ tak lagaataar girta hai. Doosre chalne wale intervals $[π, 2 π]$ , $[- π, 0]$ aur, general roop mein, $[k π, (k + 1) π]$ hain; har ek $\cos^{- 1}$ ki ek branch deta hai jiska domain $[- 1, 1]$ hai.

Inverse cosine

\begin{array}{c} \cos^{- 1} : [- 1, 1] \\ \to [0, π] \end{array}

To $y = \cos^{- 1} x$ ka matlab hai ki $\cos y = x$ aur $0 \leq y \leq π$ . Piece ko $y = x$ mein reflect karne se ek curve milta hai jo $(- 1, π)$ se $(1, 0)$ tak girta hai (Fig. 2.4). Dhyaan do ki negative number ka $\cos^{- 1}$ ek obtuse angle hota hai, kabhi negative nahi: poori range $[0, π]$ non-negative hai.

3/3 $x = - \frac{1}{2}$ se upar curve tak jao aur phir side mein: $\cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ .

Fig. 2.4

\cos^{- 1}

ki principal branch

0

π

tak values leti hai.NCERT Fig. 2.2 · p. 21

Dono axes par equal scales. Step 1 mein poora cosine curve ek halki dotted line ke roop mein dikhta hai, aur x = 0 se x = pi tak ka piece mota aur blue hai, jo 1 se minus 1 tak girta hai. Step 2 dashed mirror line y = x aur red dashed image y = cos inverse x jodta hai, jo (minus 1, pi) se (1, 0) tak girti hai. Step 3 image par x = minus one half ke upar wala point mark karta hai, dono axes tak dotted guides ke saath; yeh height 2 pi by 3 par hai.

Inverse cosecant aur secant

Cosecant sine ka reciprocal hai, isliye jo sine piece humare paas pehle se hai wahi zyada kaam kar deta hai. $[- \frac{π}{2}, \frac{π}{2}]$ par sine ek baar $[- 1, 1]$ se guzarta hai. Reciprocal lene se har non-zero value $s$ ban jaati hai $\frac{1}{s}$ , ek aisa number jiska $∣ \frac{1}{s} ∣ \geq 1$ , lekin humein $x = 0$ hatana padega, jahan sine zero hai. Jo bachta hai, $[- \frac{π}{2}, \frac{π}{2}] - {0}$ , woh one-one tarike se un saare numbers $y$ par jaata hai jinka $∣ y ∣ \geq 1$ , yaani set $𝐑 - (- 1, 1)$ , jo cosecant ki poori range hai.

Inverse cosecant

\begin{array}{c} {cosec}^{- 1} : 𝐑 - (- 1, 1) \\ \to [- \frac{π}{2}, \frac{π}{2}] - {0} \end{array}

Secant bhi isi tarah cosine piece se banta hai. $[0, π]$ par cosine ek baar $[- 1, 1]$ se guzarta hai aur $\frac{π}{2}$ par zero hai; us point ko hata kar reciprocal lene se $[0, π] - {\frac{π}{2}}$ ka $𝐑 - (- 1, 1)$ par ek one-one map milta hai.

Inverse secant

\begin{array}{c} \sec^{- 1} : 𝐑 - (- 1, 1) \\ \to [0, π] - {\frac{π}{2}} \end{array}

Har hataya gaya point inverse graph ka ek asymptote ban jaata hai. Jaise-jaise $∣ x ∣$ badhta hai, ${cosec}^{- 1} x$ dheere-dheere $0$ ki taraf aur $\sec^{- 1} x$ $\frac{π}{2}$ ki taraf jaata hai, lekin dono kabhi wahan pahunchte nahi (Fig. 2.5). Doosri branches cosecant ke liye $[\frac{π}{2}, \frac{3 π}{2}] - {π}$ aur secant ke liye $[π, 2 π] - {\frac{3 π}{2}}$ jaise sets se aati hain.

4/4 $y = \sec^{- 1} x$ $x = 1$ par $0$ se aur $x = - 1$ par $π$ se shuru hota hai; dono arms $y = \frac{π}{2}$ ke paas jaate hain.

Fig. 2.5Reflection ke baad hataye gaye points asymptotes ban jaate hain.NCERT Fig. 2.3, 2.4 · p. 22

Dono axes par lagbhag minus 3.5 se 3.5 tak equal scales. Step 1 mein y = cosec x ke do blue arms minus pi by 2 aur pi by 2 ke beech dikhte hain, ek y = 1 ke upar aur ek y = minus 1 ke neeche, jo y-axis ke paas tezi se badhte hain. Step 2 mirror line y = x aur red dashed image y = cosec inverse x jodta hai: x kam se kam 1 aur x zyada se zyada minus 1 ke liye do arms, jo heights pi by 2 aur minus pi by 2 se shuru hokar x-axis ki taraf flat hote jaate hain. Step 3 inki jagah 0 se pi par y = sec x ke do arms dikhata hai, x = pi by 2 par dashed vertical asymptote ke saath. Step 4 image y = sec inverse x jodta hai: ek arm x = 1 par 0 se badhta hai aur ek x = minus 1 par pi se girta hai, dono dashed line y = pi by 2 ke paas jaate hain.

In dono ke liye naye tables ki zaroorat kam hi padti hai. $∣ x ∣ \geq 1$ ke liye, jis angle ka secant $x$ hai, uska cosine $\frac{1}{x}$ hai, aur principal ranges bhi match karti hain, isliye $\sec^{- 1} x = \cos^{- 1} \frac{1}{x}$ . Isi tarah ${cosec}^{- 1} x = \sin^{- 1} \frac{1}{x}$ .

Inverse tangent aur cotangent

Do paas-paas ke asymptotes ke beech tangent har real number se bilkul ek baar guzarta hai. Origin ke sabse paas wala pair, $x = \pm \frac{π}{2}$ , principal piece $(- \frac{π}{2}, \frac{π}{2})$ deta hai, jis par tangent one-one aur $𝐑$ par onto hai. Isliye $\tan^{- 1}$ har real number accept karta hai, aur uski values strictly $- \frac{π}{2}$ aur $\frac{π}{2}$ ke beech rehti hain.

Inverse tangent

\tan^{- 1} : 𝐑 \to (- \frac{π}{2}, \frac{π}{2})

Cotangent ke asymptotes $π$ ke multiples par hain, aur $0$ aur $π$ ke beech yeh har real number se ek baar girte hue guzarta hai. Woh interval, $(0, π)$ , iska principal piece hai.

Inverse cotangent

\cot^{- 1} : 𝐑 \to (0, π)

Dono ranges open intervals hain. Kisi bhi real $x$ ke liye $\tan^{- 1} x = \frac{π}{2}$ nahi hota, kyunki $\tan \frac{π}{2}$ exist nahi karta. Restricted tangent aur cotangent ke vertical asymptotes unke inverses ke horizontal asymptotes ban jaate hain (Fig. 2.6). $\tan^{- 1}$ ki doosri branches $(\frac{π}{2}, \frac{3 π}{2})$ , $(- \frac{3 π}{2}, - \frac{π}{2})$ waghairah use karti hain; $\cot^{- 1}$ ki doosri branches $(π, 2 π)$ , $(- π, 0)$ waghairah use karti hain.

2/4 $y = \tan^{- 1} x$ saare $x$ ke liye defined hai aur lines $y = \pm \frac{π}{2}$ ke beech rehta hai.

Fig. 2.6Piece ke vertical asymptotes inverse ke horizontal asymptotes ban jaate hain.NCERT Fig. 2.5, 2.6 · p. 24

Lagbhag minus 3.5 se 3.5 tak equal scales. Step 1 mein blue curve y = tan x, x = minus pi by 2 aur pi by 2 par dashed vertical asymptotes ke beech dikhta hai. Step 2 mirror line y = x aur red dashed image y = tan inverse x jodta hai, jo poori picture ke across jaati hai aur y = pi by 2 aur y = minus pi by 2 par dashed horizontal lines ki taraf flat hoti jaati hai. Step 3 iske badle 0 se pi par y = cot x dikhata hai, jo x = 0 aur x = pi par asymptotes ke beech girta hai. Step 4 y = cot inverse x jodta hai, jo left mein pi ke paas se right mein 0 ke paas tak girta hai aur y-axis ko pi by 2 par kaatta hai.

Chhahon principal branches

Table 2.2 chhahon choices ko saath-saath rakhta hai. Ranges ko do families ki tarah yaad karo: $\sin^{- 1}$ , $\tan^{- 1}$ aur ${cosec}^{- 1}$ $0$ ke aas-paas, $[- \frac{π}{2}, \frac{π}{2}]$ ke andar angles dete hain; $\cos^{- 1}$ , $\cot^{- 1}$ aur $\sec^{- 1}$ $[0, π]$ ke andar angles dete hain.

Table 2.2Inverse trigonometric functions ki principal value branchesNCERT p. 25

Function	Domain	Principal range
$\sin^{- 1} x$	$[- 1, 1]$	$[- \frac{π}{2}, \frac{π}{2}]$
$\cos^{- 1} x$	$[- 1, 1]$	$[0, π]$
${cosec}^{- 1} x$	$𝐑 - (- 1, 1)$	$[- \frac{π}{2}, \frac{π}{2}] - {0}$
$\sec^{- 1} x$	$𝐑 - (- 1, 1)$	$[0, π] - {\frac{π}{2}}$
$\tan^{- 1} x$	$𝐑$	$(- \frac{π}{2}, \frac{π}{2})$
$\cot^{- 1} x$	$𝐑$	$(0, π)$

InteractiveBranch lab: chhah functions, har ek ki teen branches

Buttons se chhah trigonometric functions mein se ek aur teen branches mein se ek chuni jaati hai: principal branch, ya left ya right wali agli branch. Graph poore function ko ek halke dotted curve ke roop mein dikhata hai, chuni hui branch blue mein uske asymptotes ke saath, mirror line y = x aur, jab reflect box tick ho, red mein mirror image. Ek slider aur exact values wale buttons, jaise minus one half ya root 3, input x set karte hain; branch par point P aur image par P prime uske saath chalte hain. Ek panel chuni hui branch par angle radians mein deta hai, pi ke multiple ke roop mein jab woh ho, aur degrees mein, use principal value se compare karta hai, aur batata hai jab one half jaise input ka koi inverse secant ya cosecant nahi hota.

Fig. 2.7Ek function aur ek branch chuno, use

y = x

mein reflect karo, aur input khiska kar inverse padho.

Lab se table ko test karo. Principal branch par readout hamesha upar di gayi range mein aata hai. Paas wali branch par switch karo to same input ek alag angle deta hai jiska sine, cosine ya tangent same hai. Secant ke liye input $\frac{1}{2}$ try karo: lab mana kar deta hai, kyunki kisi angle ka secant $\frac{1}{2}$ nahi hota.

Unit circle se dono families ko imagine karna aasaan ho jaata hai (Fig. 2.8). $\sin^{- 1}$ aur $\tan^{- 1}$ ki principal values woh angles hain jo circle ke right half par khatam hote hain; $\cos^{- 1}$ aur $\cot^{- 1}$ ki values upper half par khatam hoti hain. Principal value nikaalne ke liye diye gaye ratio wale angles dhoondho aur sahi half wala rakho.

3/4Line $x = - \frac{1}{2}$ circle se $\frac{2 π}{3}$ aur $- \frac{2 π}{3}$ par milti hai; upar wala $\cos^{- 1} (- \frac{1}{2})$ hai.

Fig. 2.8

\sin^{- 1}

aur

\tan^{- 1}

ki principal values right half par khatam hoti hain;

\cos^{- 1}

aur

\cot^{- 1}

ki upper half par.

Left mein ek unit circle aur right mein notes panel. Step 1 circle ke right half ko blue mein mota karta hai, neeche minus pi by 2 se upar pi by 2 tak, aur panel sin inverse aur tan inverse ki ranges list karta hai. Step 2 upper half ko orange mein mota karta hai, right mein 0 se left mein pi tak, aur panel cos inverse aur cot inverse list karta hai. Step 3 vertical line x = minus one half banata hai, jo circle se upar 2 pi by 3 par aur neeche minus 2 pi by 3 par milti hai; upar wala point bhara hua hai aur centre se juda hai, isliye cos inverse of minus one half 2 pi by 3 hai. Step 4 horizontal line y = minus one half banata hai, jo circle se right mein minus pi by 6 par aur left mein 7 pi by 6 par milti hai; right wala point rakha jaata hai, isliye sin inverse of minus one half minus pi by 6 hai.

Definition

Principal value

Woh ek answer jo inverse trigonometric function apni principal branch fix hone ke baad deta hai: diye gaye ratio wala woh angle jo us branch mein aata hai.

Example $\sin \frac{5 π}{6} = \frac{1}{2}$ , lekin $\sin^{- 1} \frac{1}{2}$ ki principal value $\frac{π}{6}$ hai.

Example 1NCERT Example 1 · p. 26

Inverse sine ki principal value

$\sin^{- 1} (\frac{\sqrt{3}}{2})$ ki principal value nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2NCERT Example 2 · p. 26

Inverse cotangent ke liye negative input

$\cot^{- 1} (- 1)$ ki principal value nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Secant aur cosecant ke liye negative inputs

(a) $\sec^{- 1} (- \sqrt{2})$ aur (b) ${cosec}^{- 1} (- \frac{2}{\sqrt{3}})$ ki principal values nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

\tan^{- 1} (- 1)

ki principal value nikaalo. Use

k π

ke roop mein likho aur

k

type karo (jaise -1/3).

Answer dekho

Answer -0.25 π

\tan (- \frac{π}{4}) = - 1

aur

- \frac{π}{4}

(- \frac{π}{2}, \frac{π}{2})

mein aata hai, isliye

k = - \frac{1}{4}

. Answer

\frac{3 π}{4}

ka tangent bhi

- 1

hai, lekin woh principal branch ke bahar hai.

Neeche ke practice set ke Questions 1 se 12 mein answer $k π$ ko $k$ ke roop mein type karo: $- \frac{π}{3}$ ke liye -1/3 type karo, aur $π$ ke liye 1.

Practice · 14 questions

Practice 2.1: answers ke saath

Short answerEasyNCERT Exercise 2.1, Q1
$\sin^{- 1} (- \frac{\sqrt{3}}{2})$ ki principal value nikaalo.
π
Solution dikhao
$\sin (- \frac{π}{3}) = - \sin \frac{π}{3} = - \frac{\sqrt{3}}{2}$ , aur $- \frac{π}{3}$ $[- \frac{π}{2}, \frac{π}{2}]$ mein aata hai. Principal value $- \frac{π}{3}$ hai.
Short answerEasyNCERT Exercise 2.1, Q2
$\cos^{- 1} (\frac{1}{\sqrt{2}})$ ki principal value nikaalo.
π
Solution dikhao
$\cos \frac{π}{4} = \frac{1}{\sqrt{2}}$ aur $\frac{π}{4} \in [0, π]$ , isliye principal value $\frac{π}{4}$ hai.
Short answerEasyNCERT Exercise 2.1, Q3
${cosec}^{- 1} (\frac{2}{\sqrt{3}})$ ki principal value nikaalo.
π
Solution dikhao
$cosec y = \frac{2}{\sqrt{3}}$ ka matlab hai $\sin y = \frac{\sqrt{3}}{2}$ . $y \in [- \frac{π}{2}, \frac{π}{2}] - {0}$ ke saath $y = \frac{π}{3}$ .
Short answerEasyNCERT Exercise 2.1, Q4
$\tan^{- 1} (- \frac{1}{\sqrt{3}})$ ki principal value nikaalo.
π
Solution dikhao
$\tan (- \frac{π}{6}) = - \frac{1}{\sqrt{3}}$ aur $- \frac{π}{6} \in (- \frac{π}{2}, \frac{π}{2})$ , isliye principal value $- \frac{π}{6}$ hai.
Short answerEasyNCERT Exercise 2.1, Q5
$\cos^{- 1} (- \frac{\sqrt{3}}{2})$ ki principal value nikaalo.
π
Solution dikhao
$\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ , isliye $\cos (π - \frac{π}{6}) = - \frac{\sqrt{3}}{2}$ . Angle $\frac{5 π}{6}$ $[0, π]$ mein aata hai, isliye yahi principal value hai.
Short answerHardNCERT Exercise 2.1, Q6
$\tan^{- 1} (\sqrt{3} - 2)$ ki principal value nikaalo.
π
Solution dikhao
$\tan \frac{π}{12} = \frac{\sqrt{3} - 1}{1 + \sqrt{3}} = \frac{(\sqrt{3} - 1)^{2}}{2} = 2 - \sqrt{3}$ . Isliye $\tan (- \frac{π}{12}) = \sqrt{3} - 2$ , aur $- \frac{π}{12}$ $(- \frac{π}{2}, \frac{π}{2})$ mein aata hai. Principal value $- \frac{π}{12}$ hai.
Short answerEasyNCERT Exercise 2.1, Q7
$\sec^{- 1} (1)$ ki principal value nikaalo.
π
Solution dikhao
$\sec 0 = 1$ aur $0$ $[0, π] - {\frac{π}{2}}$ mein aata hai, isliye $\sec^{- 1} 1 = 0$ . ${cosec}^{- 1} 1 = \frac{π}{2}$ se compare karo: $0$ ko ${cosec}^{- 1}$ ki range chhodti hai, $\sec^{- 1}$ ki nahi.
Short answerEasyNCERT Exercise 2.1, Q8
$\cot^{- 1} (\frac{1}{\sqrt{3}})$ ki principal value nikaalo.
π
Solution dikhao
$\cot \frac{π}{3} = \frac{1}{\sqrt{3}}$ aur $\frac{π}{3} \in (0, π)$ , isliye principal value $\frac{π}{3}$ hai.
Short answerEasyNCERT Exercise 2.1, Q9
$\cos^{- 1} (- 1)$ ki principal value nikaalo.
π
Solution dikhao
$\cos π = - 1$ aur $π$ $[0, π]$ ka right-hand end hai, isliye $\cos^{- 1} (- 1) = π$ .
Short answerEasyNCERT Exercise 2.1, Q10
${cosec}^{- 1} (- 2)$ ki principal value nikaalo.
π
Solution dikhao
$cosec y = - 2$ ka matlab hai $\sin y = - \frac{1}{2}$ , aur $[- \frac{π}{2}, \frac{π}{2}] - {0}$ mein isse $y = - \frac{π}{6}$ milta hai.
Short answerMediumNCERT Exercise 2.1, Q11
$\cot^{- 1} (- 1) + \sin^{- 1} \frac{1}{2} - \sec^{- 1} \sqrt{2}$ ki value nikaalo.
π
Solution dikhao
$\cot^{- 1} (- 1) = \frac{3 π}{4}$ , $\sin^{- 1} \frac{1}{2} = \frac{π}{6}$ aur $\sec^{- 1} \sqrt{2} = \frac{π}{4}$ . Total $\frac{9 π}{12} + \frac{2 π}{12} - \frac{3 π}{12} = \frac{8 π}{12} = \frac{2 π}{3}$ hai.
Short answerMediumNCERT Exercise 2.1, Q12
$2 \cos^{- 1} \frac{\sqrt{3}}{2} + \sin^{- 1} (- 1)$ ki value nikaalo.
π
Solution dikhao
$\cos^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{6}$ aur $\sin^{- 1} (- 1) = - \frac{π}{2}$ , isliye value $\frac{π}{3} - \frac{π}{2} = - \frac{π}{6}$ hai.
Quick answerEasyNCERT Exercise 2.1, Q13
$y = \cot^{- 1} x$ ke baare mein kaun sa statement har real $x$ ke liye sahi hai?
Jab $x$ kaafi bada ho jaaye to $y = 0$ $- \frac{π}{2} < y < \frac{π}{2}$ $0 < y < π$ $x = 1$ hone par $y = \frac{π}{2}$
Answer dekho
Answer (C) $0 < y < π$
$\cot^{- 1}$ ki principal branch open interval $(0, π)$ hai: $\cot 0$ aur $\cot π$ exist nahi karte, isliye bade $x$ ke liye $y$ $0$ ke paas aata hai lekin kabhi uske barabar nahi hota. $(- \frac{π}{2}, \frac{π}{2})$ $\tan^{- 1}$ ki range hai, aur $\cot^{- 1} 1 = \frac{π}{4}$ .
Quick answerMediumNCERT Exercise 2.1, Q14
$\sec^{- 1} (- \sqrt{2}) - \tan^{- 1} (- \frac{1}{\sqrt{3}})$ kiske barabar hai?
$\frac{7 π}{12}$ $\frac{11 π}{12}$ $- \frac{π}{12}$ $- \frac{5 π}{12}$
Answer dekho
Answer (B) $\frac{11 π}{12}$
$\sec^{- 1} (- \sqrt{2}) = \frac{3 π}{4}$ aur $\tan^{- 1} (- \frac{1}{\sqrt{3}}) = - \frac{π}{6}$ , isliye value $\frac{3 π}{4} + \frac{π}{6} = \frac{11 π}{12}$ hai. $\sec^{- 1} (- \sqrt{2}) = - \frac{π}{4}$ lene par, jo uski range ke bahar hai, $- \frac{π}{12}$ aata hai.

2.3 Inverse Trigonometric Functions ki Properties

Is section ki har identity ke saath fine print hai. Woh sirf un $x$ ke liye sahi hai jahan dono sides defined hon aur jo angles aate hain woh principal branches ke andar rahein. Fine print ko ignore karna is chapter mein galat answers ki sabse common wajah hai, isliye neeche har result ke saath uska interval diya gaya hai.

PropertyTrigonometric function ko undo karna

$x \in [- 1, 1]$ ke liye $\sin (\sin^{- 1} x) = x$ , aur $x \in [- \frac{π}{2}, \frac{π}{2}]$ ke liye $\sin^{- 1} (\sin x) = x$ . Isi tarah $x \in [- 1, 1]$ ke liye $\cos (\cos^{- 1} x) = x$ aur $x \in [0, π]$ ke liye $\cos^{- 1} (\cos x) = x$ ; har real $x$ ke liye $\tan (\tan^{- 1} x) = x$ aur $x \in (- \frac{π}{2}, \frac{π}{2})$ ke liye $\tan^{- 1} (\tan x) = x$ . Baaki teen functions apni-apni branches par isi pattern ko follow karte hain.

Dono halves ek jaise nahi hain. Pehla, $\sin (\sin^{- 1} x) = x$ , $\sin^{- 1}$ ke poore domain par sahi hai. Doosra tab toot jaata hai jaise hi $x$ $[- \frac{π}{2}, \frac{π}{2}]$ se bahar jaata hai: jaise $\sin^{- 1} (\sin π) = \sin^{- 1} 0 = 0$ , $π$ nahi. $y = \sin^{- 1} (\sin x)$ ka graph dikhata hai ki iske badle kya hota hai (Fig. 2.9).

3/3 $x = \frac{5 π}{4}$ par graph $π - \frac{5 π}{4} = - \frac{π}{4}$ deta hai.

Fig. 2.9

\sin^{- 1} (\sin x)

sirf principal branch par

x

ke barabar hai; baaki jagah yeh wapas mud jaata hai.

Equal scales, x lagbhag minus pi se 2 pi tak. Step 1 minus pi by 2 se pi by 2 tak ki strip shade karta hai, jahan ek chhota blue segment bilkul dashed line y = x par hai. Step 2 baaki blue graph jodta hai: ek zigzag jiska slope pi by 2 se 3 pi by 2 par minus pi by 2 tak minus 1 hai, phir slope dobara 1, aur left mein bhi waisa hi mod, jo kabhi pi by 2 se upar ya minus pi by 2 se neeche nahi jaata. Step 3 x = 5 pi by 4 ke upar wala point mark karta hai, jo height minus pi by 4 par hai, dono axes tak dotted guides ke saath.

Zigzag ka slope $[- \frac{π}{2}, \frac{π}{2}]$ par $1$ , $[\frac{π}{2}, \frac{3 π}{2}]$ par $- 1$ , $[\frac{3 π}{2}, \frac{5 π}{2}]$ par phir se $1$ hai, aur yeh period $2 π$ ke saath repeat hota hai. Kisi bhi $a$ ke liye $\sin^{- 1} (\sin a)$ evaluate karne ke liye, $\sin (π - a) = \sin a$ aur $\sin (a + 2 π) = \sin a$ use karke $a$ ki jagah $[- \frac{π}{2}, \frac{π}{2}]$ ka same sine wala angle rakho. Yahi idea baaki functions ke liye unki apni symmetries ke saath kaam karta hai, jaise $\cos (2 π - a) = \cos a$ aur $\tan (a + π) = \tan a$ .

Zyaadatar simplifications ek hi recipe follow karte hain. $x$ ko kisi naye angle $θ$ ka trigonometric function maan lo, expression ko standard identity se dobara likho, aur upar wali property se inverse cancel karo. Dhyaan aakhri step mein chahiye: cancel karne se pehle check karo ki jis angle tak tum pahunche ho woh sahi principal branch mein hai.

Example 3NCERT Example 3 · p. 28

Ek expression, do answers

Dikhao ki $0 \leq x \leq 1$ ke liye $\cos^{- 1} (2 x^{2} - 1) = 2 \cos^{- 1} x$ , aur $- 1 \leq x \leq 0$ ke liye $\cos^{- 1} (2 x^{2} - 1) = 2 π - 2 \cos^{- 1} x$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Fig. 2.10 Example 3 ki dono sides ko compare karta hai. $[0, 1]$ par woh ek ho jaati hain. $[- 1, 0)$ par curve $y = 2 \cos^{- 1} x$ $π$ se upar chala jaata hai, jo height $\cos^{- 1}$ ki koi value reach nahi kar sakti, isliye left-hand side ko wapas mudna padta hai.

3/3 $[0, 1]$ par dono match karte hain. $x = - \frac{1}{2}$ par woh $\frac{4 π}{3}$ aur $\frac{2 π}{3}$ dete hain, aur $\frac{2 π}{3} = 2 π - \frac{4 π}{3}$ .

Fig. 2.10Example 3 ki dono sides sirf

0 \leq x \leq 1

ke liye match karti hain.

x-axis minus 1 se 1 tak aur y-axis 0 se 2 pi tak hai. Step 1 blue curve y = 2 cos inverse x dikhata hai, jo x = minus 1 par 2 pi se x = 1 par 0 tak girta hai. Step 2 orange dashed curve y = cos inverse of (2 x squared minus 1) jodta hai, jo x = minus 1 par 0 se x = 0 par pi tak badhta hai aur wapas 0 tak girta hai, y-axis ke around symmetric. Step 3 0 se 1 tak ka hissa shade karta hai, jis par likha hai yahan equal, jahan dono curves ek ho jaate hain, aur x = minus one half mark karta hai, jahan blue curve 4 pi by 3 par aur orange wala 2 pi by 3 par hai.

Example 4NCERT Example 4 · p. 28

Inverse tangent ke andar sum-to-product

$\tan^{- 1} (\frac{\sin 3 x + \sin x}{\cos 3 x + \cos x})$ , $- \frac{π}{4} < x < \frac{π}{4}$ , ko simplest form mein likho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5NCERT Example 5 · p. 29

Ek surd jismein angle chhupa hai

$\sin^{- 1} (\frac{1}{\sqrt{1 + x^{2}}})$ , $x > 0$ , ko simplest form mein likho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Miscellaneous examples

Example 6NCERT Example 6 · p. 30

Branch ke bahar ka angle

$\sin^{- 1} (\sin \frac{5 π}{4})$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

$\cos (\tan^{- 1} x)$ ya $\tan (\sin^{- 1} \frac{7}{25})$ jaise compositions right triangle se solve hote hain. Andar wale angle ko $θ$ naam do, ek triangle banao jiski sides diye gaye ratio mein fit hon, Pythagoras se teesri side nikaalo aur jo ratio chahiye woh padh lo (Fig. 2.11). Negative input ke liye triangle answer ka size deta hai, aur $θ$ ka quadrant uska sign.

3/3Ab $\cos θ = \frac{24}{25}$ aur $\tan θ = \frac{7}{24}$ , isliye $\tan (\sin^{- 1} \frac{7}{25}) = \frac{7}{24}$ .

Fig. 2.11Ratio se do sides, Pythagoras se teesri, phir jo ratio chaho.

Ek right triangle jiska angle theta left corner par aur right angle neeche right mein hai. Upar ek panel kehta hai theta equals sin inverse of 7 by 25. Step 1 theta ke saamne wali chhoti vertical side par 7 aur tirchhe hypotenuse par 25 likhta hai, aur base par question mark lagata hai. Step 2 base ko orange karta hai aur us par 24 likhta hai, aur panel dikhata hai square root of 25 squared minus 7 squared equals 24. Step 3 sides ko adjacent aur opposite naam deta hai, aur panel par likha hai cos theta equals 24 by 25 aur tan theta equals 7 by 24.

Example

Quadrant par nazar rakho

(a) $\tan (\cos^{- 1} (- \frac{3}{5}))$ aur (b) $\cos (\tan^{- 1} (- 2))$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

\cos^{- 1} (\cos \frac{7 π}{4})

nikaalo. Use

k π

ke roop mein likho aur

k

type karo.

Answer dekho

Answer 0.25 π

\cos \frac{7 π}{4} = \cos (2 π - \frac{7 π}{4}) = \cos \frac{π}{4}

, aur

\frac{π}{4} \in [0, π]

. Isliye value

\frac{π}{4}

hai aur

k = \frac{1}{4}

History seNCERT Historical Note · p. 33

Jyā se arc sine tak

Sine function Indian astronomy se nikla. Siddhānta astronomers circle ke half-chord ke saath kaam karte the, jise woh jyā ya ardha-jyā kehte the, na ki Greek astronomers wale poore chord ke saath. Āryabhaṭa (janm 476 CE) ne apne Āryabhaṭīya (499 CE) mein $3 {\frac{3}{4}}^{\circ}$ ke steps par in half-chords ka table diya. Yeh word Arabic mein jība bana, baad mein ise jaib (fold ya pocket) padha gaya aur yeh Latin sinus ban gaya: hamara “sine”.

Baad ke astronomers ne in tools ko aur behtar banaya. Bhāskara I (7th century) ne ek simple fraction dhoondha jo kisi angle ke sine ko kamaal ki accuracy se approximate karta hai, Brahmagupta ne sine table ki entries ke beech ki values padhne ke rules diye, aur Bhāskara II (janm 1114 CE) ne $\sin 18^{\circ}$ aur $\sin 36^{\circ}$ jaisi exact values nikaali.

Inverse ke pehle ideas Kerala se aaye. Saṅgamagrāma ke Mādhava (lagbhag 1340 se 1425) ne kisi arc ke liye uske tangent ke terms mein ek kabhi na khatam hone wali series dhoondhi, jise hum ab $\tan^{- 1} x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \dots$ likhte hain, aur usse $π$ ki ek series bhi. Malayalam text Yuktibhāṣā (lagbhag 1530) ne in results ke proofs diye, saath mein $\sin (A + B)$ ka expansion bhi.

Notation $\sin^{- 1} x$ English astronomer John Herschel ne 1813 mein propose kiya; purana naam arc sine yeh batata hai ki answer unit circle ka ek arc hai. Isse bahut pehle, Miletus ke Thales (lagbhag 600 BCE) ne kehte hain ek pyramid ki height uski parchhai ko ek seedhe khade dande ki parchhai se compare karke nikaali thi, yeh us ratio ka shuruaati use tha jise hum ab sun ki altitude ka tangent kehte hain. Āryabhaṭīya mein bhi isi tarah ke shadow problems hain, lamp ke paas khade gnomon (śaṅku) ke baare mein.

Timeline

Inverse trigonometric functions tak ke safar ke milestones

499 CE
Āryabhaṭīya
$3 {\frac{3}{4}}^{\circ}$ ke steps par half-chords (jyā) ka table.
c. 629 CE
Bhāskara I
Sine ko approximate karne wala ek simple rational formula.
665 CE
Brahmagupta
Sine tables ke liye second-order interpolation.
1150 CE
Bhāskara II
$\sin 18^{\circ}$ aur $\sin 36^{\circ}$ jaisi exact values.
c. 1400
Mādhava
Arc tangent, $\tan^{- 1} x$ , aur $π$ ki series.
c. 1530
Yuktibhāṣā
Kerala series ke proofs, Malayalam mein likhe gaye.
1813
John Herschel
Arc sine ke liye notation $\sin^{- 1} x$ introduce karte hain.

Flash cards · 8

Principal branches: jaldi se yaad karo

1 / 8

Recap

Summary aur key points

Trigonometric function apne natural domain par many-one hota hai, isliye uska inverse tabhi hota hai jab uske domain ko aise interval par restrict kiya jaaye jahan woh one-one aur onto ho (Section 2.1).
Agar $f$ one-one aur onto hai, to $f^{- 1} (f (x)) = x$ aur $f (f^{- 1} (y)) = y$ , aur $f^{- 1}$ ka graph line $y = x$ mein $f$ ke graph ki mirror image hai (Section 2.2).
Har chalne wala interval inverse ki ek branch deta hai. Tay ki gayi branch principal value branch hai, aur usse li gayi value principal value hai (Inverse sine).
$\sin^{- 1} : [- 1, 1] \to [- \frac{π}{2}, \frac{π}{2}]$ aur $\cos^{- 1} : [- 1, 1] \to [0, π]$ (Inverse cosine).
${cosec}^{- 1}$ aur $\sec^{- 1}$ $∣ x ∣ \geq 1$ ke liye defined hain, ranges $[- \frac{π}{2}, \frac{π}{2}] - {0}$ aur $[0, π] - {\frac{π}{2}}$ ke saath (Inverse cosecant aur secant).
$\tan^{- 1}$ aur $\cot^{- 1}$ har real $x$ ke liye defined hain, open ranges $(- \frac{π}{2}, \frac{π}{2})$ aur $(0, π)$ ke saath (Inverse tangent aur cotangent).
$\sin^{- 1} x$ ek angle hai, jabki $(\sin x)^{- 1} = cosec x$ ek reciprocal hai. Jab koi branch naam se na batayi jaaye, to principal branch hi samjho (Chhahon principal branches).
Negative input ke liye $\sin^{- 1}$ , $\tan^{- 1}$ aur ${cosec}^{- 1}$ negative angle dete hain, jabki $\cos^{- 1}$ , $\cot^{- 1}$ aur $\sec^{- 1}$ $\frac{π}{2}$ aur $π$ ke beech ka angle dete hain (negative inputs).
$[- 1, 1]$ par $\sin (\sin^{- 1} x) = x$ , lekin $\sin^{- 1} (\sin x) = x$ sirf $[- \frac{π}{2}, \frac{π}{2}]$ par; baaki har function apni branch par yahi do rules follow karta hai (Property).
Branch ke bahar wale $a$ ke liye $\sin^{- 1} (\sin a)$ nikaalne ke liye, $a$ ki jagah $[- \frac{π}{2}, \frac{π}{2}]$ ka same sine wala angle rakho (Example 6).
$x = \sin θ$ , $\cos θ$ ya $\tan θ$ jaise substitutions inverse trigonometric expressions ko simplify karte hain; $θ$ ka interval final form tay karta hai (Example 3).
$\tan (\cos^{- 1} x)$ jaise compositions right triangle se padhe jaate hain, aur sign principal value ke quadrant se tay hota hai (Fig. 2.11).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 2.1, Q1	Practice 2.1 · Q1
NCERT Exercise 2.1, Q2	Practice 2.1 · Q2
NCERT Exercise 2.1, Q3	Practice 2.1 · Q3
NCERT Exercise 2.1, Q4	Practice 2.1 · Q4
NCERT Exercise 2.1, Q5	Practice 2.1 · Q5
NCERT Exercise 2.1, Q6	Practice 2.1 · Q6
NCERT Exercise 2.1, Q7	Practice 2.1 · Q7
NCERT Exercise 2.1, Q8	Practice 2.1 · Q8
NCERT Exercise 2.1, Q9	Practice 2.1 · Q9
NCERT Exercise 2.1, Q10	Practice 2.1 · Q10
NCERT Exercise 2.1, Q11	Practice 2.1 · Q11
NCERT Exercise 2.1, Q12	Practice 2.1 · Q12
NCERT Exercise 2.1, Q13	Practice 2.1 · Q13
NCERT Exercise 2.1, Q14	Practice 2.1 · Q14
NCERT Exercise 2.2, Q1	Practice 2.2 · Q1
NCERT Exercise 2.2, Q2	Practice 2.2 · Q2
NCERT Exercise 2.2, Q3	Practice 2.2 · Q3
NCERT Exercise 2.2, Q4	Practice 2.2 · Q4
NCERT Exercise 2.2, Q5	Practice 2.2 · Q5
NCERT Exercise 2.2, Q6	Practice 2.2 · Q6
NCERT Exercise 2.2, Q7	Practice 2.2 · Q7
NCERT Exercise 2.2, Q8	Practice 2.2 · Q8
NCERT Exercise 2.2, Q9	Practice 2.2 · Q9
NCERT Exercise 2.2, Q10	Practice 2.2 · Q10
NCERT Exercise 2.2, Q11	Practice 2.2 · Q11
NCERT Exercise 2.2, Q12	Practice 2.2 · Q12
NCERT Exercise 2.2, Q13	Practice 2.2 · Q13
NCERT Exercise 2.2, Q14	Practice 2.2 · Q14
NCERT Exercise 2.2, Q15	Practice 2.2 · Q15
NCERT Miscellaneous Exercise, Q1	Chapter 2 par miscellaneous exercises · Q1
NCERT Miscellaneous Exercise, Q2	Chapter 2 par miscellaneous exercises · Q2
NCERT Miscellaneous Exercise, Q3	Chapter 2 par miscellaneous exercises · Q3
NCERT Miscellaneous Exercise, Q4	Chapter 2 par miscellaneous exercises · Q4
NCERT Miscellaneous Exercise, Q5	Chapter 2 par miscellaneous exercises · Q5
NCERT Miscellaneous Exercise, Q6	Chapter 2 par miscellaneous exercises · Q6
NCERT Miscellaneous Exercise, Q7	Chapter 2 par miscellaneous exercises · Q7
NCERT Miscellaneous Exercise, Q8	Chapter 2 par miscellaneous exercises · Q8
NCERT Miscellaneous Exercise, Q9	Chapter 2 par miscellaneous exercises · Q9
NCERT Miscellaneous Exercise, Q10	Chapter 2 par miscellaneous exercises · Q10
NCERT Miscellaneous Exercise, Q11	Chapter 2 par miscellaneous exercises · Q11
NCERT Miscellaneous Exercise, Q12	Chapter 2 par miscellaneous exercises · Q12
NCERT Miscellaneous Exercise, Q13	Chapter 2 par miscellaneous exercises · Q13
NCERT Miscellaneous Exercise, Q14	Chapter 2 par miscellaneous exercises · Q14

Quick check: MCQs

2.1 Introduction2 questions
Inverse cosine3 questions
Chhahon principal branches4 questions
2.3 Inverse Trigonometric Functions ki Properties4 questions
Miscellaneous examples3 questions

Glossary · 9

Key terms (English + Hindi)

asymptoteअनंतस्पर्शीanantasparshi: Aisi line jiske graph kareeb, aur kareeb aata jaata hai lekin use kabhi chhoota nahi, jaise $y = \tan^{- 1} x$ ke liye $y = \frac{π}{2}$ .Chapter mein pehli baar yahan aaya →

branchशाखाshaakhaa: Ek interval jis par trigonometric function one-one aur apni range par onto ho, ya usse milne wala inverse.Chapter mein pehli baar yahan aaya →

inverse functionप्रतिलोम फलनpratilom phalan: Bijection $f : X \to Y$ ke liye woh function $f^{- 1} : Y \to X$ jo har $f (x)$ ko wapas $x$ par bhejta hai.Chapter mein pehli baar yahan aaya →
inverse trigonometric functionप्रतिलोम त्रिकोणमितीय फलनpratilom trikonmitiya phalan: Ek branch par restrict kiye gaye trigonometric function ka inverse; jaise $\sin^{- 1} x$ $[- \frac{π}{2}, \frac{π}{2}]$ ka woh angle hai jiska sine $x$ hai.Chapter mein pehli baar yahan aaya →

many-one functionबहुएक फलनbahu-ek phalan: Aisa function jo do ya zyada alag inputs ko same output par bhejta hai, isliye woh one-one nahi hota.Chapter mein pehli baar yahan aaya →

one-one aur onto functionएकैकी आच्छादक फलनone-one and onto functionekaiki aachchhaadak phalan: Aisa function jo one-one aur onto dono ho; sirf aise functions ke hi inverses hote hain.Chapter mein pehli baar yahan aaya →

principal valueमुख्य मानmukhya maan: Woh angle jo inverse trigonometric function apni principal value branch se deta hai, jaise $\sin^{- 1} \frac{1}{2}$ ke liye $\frac{π}{6}$ .Chapter mein pehli baar yahan aaya →
principal value branchमुख्य मान शाखाmukhya maan shaakhaa: Woh branch jo convention se kisi inverse trigonometric function ki range ke roop mein fix ki gayi hai, jaise $\cos^{- 1}$ ke liye $[0, π]$ .Chapter mein pehli baar yahan aaya →

restricted domainप्रतिबंधित प्रांतpratibandhit praant: Function ke natural domain ka ek hissa, jo is tarah chuna jaata hai ki function one-one aur apni range par onto ban jaaye.Chapter mein pehli baar yahan aaya →

Inverse Trigonometric Functions: Class 12 Maths Chapter 2 Hinglish mein

Is chapter mein aap kya seekhenge

2.1 Introduction

2.2 Basic Concepts

Inverse sine

Inverse cosine

Inverse cosecant aur secant

Inverse tangent aur cotangent

Chhahon principal branches

Practice 2.1: answers ke saath

2.3 Inverse Trigonometric Functions ki Properties

Practice 2.2: answers ke saath

Miscellaneous examples

Summary aur key points

Practice questions answers ke saath (NCERT “Chapter 2 par miscellaneous exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye