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NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions

NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions

Edited By Komal Miglani | Updated on Mar 31, 2025 03:20 AM IST | #CBSE Class 12th

Imagine trying to find an unknown angle when only its sine or cosine value is given. Have you ever wondered how calculators determine the angles for trigonometric values? This is where inverse trigonometric functions come into play. They help us reverse the process of basic trigonometric functions like sine, cosine, and tangent, allowing us to find angles from given ratios. When the trigonometric values are known, the inverse trigonometric functions allow you to find the angles. These functions are very important when it comes to calculus, solving equations, and even putting them into practice in the real world, such as physics and engineering. We will go through all the NCERT Exemplar questions and solutions in this chapter, which are aimed at developing a solid understanding of this topic to help students understand what the author is trying to assess in the exam.

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  1. NCERT Exemplar Class 12 Maths Solutions Chapter 2
  2. Sub-Topics Covered in NCERT Exemplar Class 12 Maths Solutions Chapter 2
  3. NCERT Exemplar Class 12 Maths Solutions
  4. Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 2
  5. NCERT solutions of class 12 - Subject-wise
  6. NCERT Notes of class 12 - Subject Wise
  7. NCERT Books and NCERT Syllabus
  8. NCERT Exemplar Class 12 Solutions - Subject Wise

In NCERT Solutions for Class 12 Maths Chapter 2, students will understand how to find the ranges and domains of inverse trigonometric functions. Some other key concepts that students will learn are the behavior of the function, properties of inverse trigonometric functions, and more, all to be explained in a systematic way. The NCERT Exemplar Class 12 Maths Chapter 2 Solutions will provide step-by-step guidance to ensure a strong conceptual understanding for students.

NCERT Exemplar Class 12 Maths Solutions Chapter 2

Class 12 Maths Chapter 2 exemplar solutions Exercise: 2.3
Page number: 35-41
Total questions: 55
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Question:1

find the value of tan1(tan5π6)+cos1(cos13π6)

Answer:

we know that
tan1(tanx)=x;xϵ(π2,π2) and
cos1(cosx)=x;xϵ(0,π)

tan1(tan5π6)+cos1(cos13π6)
=tan1[tan(ππ6)]+cos1[cos(π+7π6)]
=tan1(tanπ6)+cos1(cos7π6) [since , cos(π+θ)=cosθ ]
=tan1(tanπ6)+π[cos1(cos7π6)]
[sincetan1(x)=tan1x,xϵRandcos1=(x)=πcos1(x),xϵ(1,1)]
=tan1(tanπ6)+πcos1[cos(π+π6)]
=tan1(tanπ6)+πcos1(cosπ6) [since , cos(π+θ)=cosθ ]
=tan1(tanπ6)+ππ+cos1(cosπ6)
=π6+0+π6
=0

Question:2

Evaluate
cos[cos1(32)+π6]

Answer:

We have
cos[cos1(32)+π6]
[Since,cos5π6=32]
=cos[cos1(cos5π6)+π6]
=cos(5π6+π6)

[since.cos1(cosx)=x;xϵ(0,π)]
=cos(6π6)
=cosπ
=-1

Question:3

Prove that cot(π42cot13)=7

Answer:

We prove that
cot(π42cot13)=7
cot(π42cot13)=cot17
2cot13=π4cot17
2tan113=π4tan117
2tan113+tan117=π4
[since,2tan1(x)=2tan12x1(x)2]
tan1231(13)2+tan117=π4
tan1(2389)+tan117=π4
tan1(34)+tan117=π4
[since,tan1x+tan1y=tan1x+y1xy]
tan1(34+17134,17)=π4
tan1(21+4)28(283)28=π4
tan12525=π4
tan1(1)=π4
1=tanπ4
1=1
LHS=RHS
Hence Proved

Question:4

Find the value of tan1(13)+cot1(13)+tan1[sin(π2)]

Answer:

We have
tan1(13)+cot1(13)+tan1[sin(π2)]
=tan1(tan5π6)+cot1(cotπ3)+tan1(1)
=tan1[tan(π5π6)]+cot1(cotπ3)+tan1[tan(ππ4)]
[since,tan1(tanx)=x,xϵ(π2,π2);cot1(cotx)=x,xϵ(0,π);andtan1(x)=tan1x]
=tan1(tanπ6)+cot1(cotπ3)+tan1[tanπ4]
=π6+π3π4
=2π+4π3π12
=π12

Question:5

find the value of tan1(tan2π3)

Answer:

We have
tan1(tan2π3)=tan1tan(ππ3)
=tan1(tanπ3)
[Since,tan1(x)=tan1x,xϵR]
=tan1tanπ3
[Since,tan1(tanx)=x,xϵ(π2,π2)]
=π3

Question:6

show that 2tan1(3)=π2+tan1(43)

Answer:

We have to prove ,
2tan1(3)=π2+tan1(43)
LHS=2tan1(3) [Since,tan1(x)=tan1x,xϵR]
=[cos11321+32] [Since2tan1x=[cos11x21+x2],x0]
=[cos1(810)]
=[cos1(45)]
=[πcos1(45)] =[sincecos1(x)=πcos1x,xϵ[1,1]]
=π+cos1(45)
[letcos1(45)=0cosθ=(45)tanθ=(34)θ=tan1(34)]
=π+tan1(34)=π+[π2cot1(34)]
=π2cot1(34)
[Since,tan1(x)=tan1x]
=π2+tan1(43)
=π2+tan1(43)
=RHS
Hence Proved.

Question:7

Find the real solution of the equation:
tan1x(x+1)+sin1x2+x+1=π2

Answer:

We have , tan1x(x+1)+sin1x2+x+1=π2.........(i)
Let sin1x2+x+1=θ
sinθ=x2+x+11
tanθ=x2+x+1x2x [Since,tanθ=sinθcosθ]
θ=tan1x2+x+1x2x=sin1x2+x+1
On Putting the value of θ in Eq. (i), We get
tan1x(x+1)+tan1x2+x+1x2x=π2.........(ii)
we know that,
tan1x+tan1y=tan1x+y1xy,xy<1
So,(ii) becomes,
tan1[x(x+1)+x2+x+1x2x1x(x+1)x2+x+1x2x]=π2
tan1[x(x+1)+x2+x+11(x2+x)1x(x+1)x2+x+11(x2+x)]=π2
x2+x+(x2+x+1)[1(x2+x+1).(x2+x)]=tanπ2=10
[1(x2+x+1).(x2+x)]=0
(x2+x+1)=1orx2+x=0
x2x1=1orx(x+1)=0
x2+x+2=0orx(x+1)=0
x=1±1(4×2)2orx=1
x=0orx=1
For real solution , we have x=0,-1.

Question:8

Find the value of sin(2tan113)+cos(tan122)

Answer:

We have sin(2tan113)+cos(tan122)
=sin[sin1{2×131+(13)2}]+cos(cos113)
[Since,tan1x=cos111+x2;2tan1(x)=2tan12x1(x)2,1x1andtan122=cos113]
=sin[sin1{231+19}]+13 [Since,cos(cos1x)=x,xϵ{1,1}]
=sin[sin1(2×93×10)]+13
=sin[sin1(35)]+13
=35+13[Since,sin(sin1x)=x]
=1415

Question:9

If 2tan1(cosθ)=tan1(cosecθ), then show that θ=π4, where n is any integer.

Answer:

We have 2tan1(cosθ)=tan1(cosecθ)
tan1(2cosθ1cos2θ)=tan1(2cosecθ) [Since2tan1x=tan1(2x1x2)]
2cosθsin2θ=2cosecθ
cotθ.2cosecθ=2cosecθ
cotθ=1
cotθ=cotπ4
θ=π4
Hence Proved

Question:10

Show that cos(2tan117)=sin(4tan113)

Answer:

We have , cos(2tan117)=sin(4tan113)
cos[cos1(1(17)21+(17)2)]=sin(2.2tan113)
[Since,2tan1x=[cos11x21+x2],x0]
cos[cos1(48495049)]=sin[2(tan1231(13)2)]
cos[cos1(2425)]=sin(2tan134)
cos[cos1(2425)]=sin(sin12×341+916) [Since,2tan1x=tan1(2x1x2)]
2425=sin(sin1322516)
2425=4850
2425=2425
Since LHS=RHS
Hence Proved

Question:11

Solve the following equation cos(tan1x)=sin(cot134)

Answer:

We have cos(tan1x)=sin(cot134)
cos(cos11x2+2)=sin(sin145)........(i)
Let tan1x=θ1tanθ1=x1
cosθ1=1x2+1.....(a)
θ1=cos11x2+1.....(c)
And cot1=θ2cot1=34
sinθ2=45.......(b)
θ2=sin145.......(d)
From (c),(d);(i) becomes
cosθ1=sinθ2
1x2+1=45 [From (a),(b)]
On squarinting both Sides, we get
16(x2+1)=25
16x2=9
x2=(34)2
x=±34
x=34,34.

Question:12

Prove that tan1(1+x2+1x21+x21x2)=π4+12cos1x2

Answer:

We have tan1(1+x2+1x21+x21x2)=π4+12cos1x2
LHS=tan1(1+x2+1x21+x21x2)........(i)
[x2=cos2θ=cos2θ+sin2θ=12sin2θ=2cos2θ1]
cos1x2=2θ
θ=12cos1x2
1+x2=1+cos2θ
1+2cos2θ1=2cosθ
And 1x2=1cos2θ
11+2sin2θ=2sinθ
LHS=tan1(2cosθ+2sinθ2cosθ2sinθ)
=tan1(cosθ+sinθcosθsinθ)
=tan1(1+tanθ1tanθ)
=tan1{tan(π4)+tanθtan(π4)tanθ}
=tan1[tan(π4+θ)] [Since,tan(x+y)=tanx+tany1tanx.tany]
=π4+θ
=π4+12cos1x2
=RHS
LHS=RHS
Hence Proved

Question:13

Find the simplified from of cos1(35cosx+45sinx) , where xϵ[3π4,π4]

Answer:

Let cosy=35
siny=45
y=cos135=sin145=tan143
cos1[cosy.cosx+siny.sinx]
[since,cos(AB)=cosA.cosB+sinA.sinB]
=cos1[cos(yx)]
[scine,cos(cos1x)=x,xϵ{1,1}]
=y-x
[scine,y=tan143]
=tan143x

Question:14

Prove that sin1817+sin135=sin17785

Answer:

we have sin1817+sin135=sin17785
LHS=sin1817+sin135

letsin1817=θ1
sinθ1=817
tanθ1=815θ1=tan1815
And, sin35=θ2sin135
tanθ2=34θ2=tan134
=tan1815+tan134
=tan1[815+341815×34] [Since,tan1x+tan1y=tan1(x+y1xy)]
=tan1[77603660]
=tan1(7736)
Let =θ3=tan1(7736)tanθ3=7736
sinθ3=775929+1296=7785
θ3=sin1(7785)
=sin1(7785)=RHS
Hence proved

Question:15

Show that sin1513+cos135=tan16316

Answer:

Solving LHS, sin1513+cos135
Letsin1513=x
sinx=513
Andcos2x=1sin2x
125169=144169
cosx=144169=1213
tanx=sinxcosx=5131213=512
tanx=512..........(i)
Again , let cos135=y
cosy=35
siny=1cos2y
siny=1(35)2
siny=1625=45
tany=sinycosy=4535=43.........(ii)
We know that, tan(x+y)=512+43151243 [from (i),(ii)]
tan(x+y)=15+4836362036
tan(x+y)=63361636
(x+y)=tan16316(x+y)=tan16316=RHS
Since , LHS=RHS
Hence Proved.

Question:16

Prove that , tan114+tan129=sin115

Answer:

Solving LHS, tan114+tan129
Let tan114=x
tanx=14
Squaring both sides,
tan2x=116
sec2x1=116
sec2x=1716
1cos2x=1716
cos2x=1617
cosx=417
Since,sin2x=1cos2x
sin2x=11617=117
sinx=117
Again,
Let tan129=y
tany=29
Squaring both sides,
tan2y=481
sec2y1=481
sec2y=8581
1cos2y=8581
cos2y=8185
cosy=985
Since, sin2y=1cos2y
sin2=18185=485
sinx=285
We know that, sin(x+y)=sinx.siny+cosx.siny
sin(x+y)=117.985+417.285
sin(x+y)=1717.85
sin(x+y)=1717.5
sin(x+y)=15
x+y=sin115=RHS
Since , LHS=RHS
Hence Proved

Question:17

Find the value of 4tan115tan11239

Answer:

We have, 4tan115tan11239
=2×(2tan115)tan11239
=2[tan1251(15)2]tan11239 [since,2tan1x=tan12x1(x)2]

=2[tan1252425]tan11239
=2tan1512tan11239
=[tan1561(512)2]tan11239 [since,2tan1x=tan12x1(x)2]
=tan156125144tan11139
=tan1(144×5119×6)tan11239
=tan1120119tan11239
=tan112011912391+120119.1239[since,tan1xtan1y=tan1(xy1+xy)]
=tan1[2868011928441+120]
=tan12856128561
=tan1(1)
=tan1(tanπ4)
=π4
Hence, 4tan115tan11239=π4

Question:18

Show that tan(12sin134)=473 and, justify why the other value 4+73 is ignored.

Answer:

Solving LHS,
=tan(12sin134)
Let12sin134=θ
sin134=2θ
34=sin2θ
sin2θ=34
2tanθ1+tan2θ=34
3+3tan2θ=8tanθ
3tan2θ8tanθ=3
Lettanθ=y
3y2+8y+3=0
y=8±644×3×32×3⇒=8±286
y=2(4±7)2×3
tanθ=(4±7)3
θ=tan1(4±7)3
{ but as we can see , 4+73>1, since max[tan(12sin134)]=1}
tan(12sin134)=473=RHS
Note: Scince π2sin134π2
π412sin134π4
tan(π4)tan(12sin134)tan(π4)
1tan(12sin134)1

Question:19

If a1,a2,a3,.................an is an arithmetic progression with common difference d, then evaluate the following expression.
tan[tan1(d1+a1a2)+tan1(d1+a2a3)+tan1(d1+a3a4)+..........+tan1(d1+an1an)]

Answer:

We have a1=a,a2=a+d,a3=a+2d.......
And, d=a2a1=a3a2=a4a3=......=anan1
Given that,
tan[tan1(d1+a1a2)+tan1(d1+a2a3)+tan1(d1+a3a4)+............+tan1(d1+an1an)]
=tan1[tan1(a2a11+a1a2)+tan1(a3a21+a2a3)+.........+tan1(anan11+an1an)]
=tan[(tan1a2tan1a1)+(tan1a3tan1a2)+................+(tan1antan1an1)]
=tan[tan1antan1a1]
[Scince,tan1xtan1y=tan1(xy1+xy)]
=tan[tan1(ana11+a1an)]
[scince,tan(tan1x)=x]
=ana11+a1an

Question:20

Which of the following in the principal value branch of cos1x
A.[π2,π2]
B.(0,π)
C.[0,π]
D.(0,π)π2

Answer:

Answer : (c)
We know that the principal value branch of cos1 is [0,π]

Question:21

Which of the following in the principal value branch of cosec1x .
A.(π2,π2)
B.[0,π]{π2}
C.[π2,π2]
D.[π2,π2]{0}

Answer:

Answer :(D)

We know that the principal value branch of cosec1x is [π2,π2](0)

Question:22

If 3tan1x+cot1x=π, then x equals to
A. 0
B. 1
C. -1
D. 1/2

Answer:

Answer : B
Given That, 3tan1x+cot1x=π
2tan1x+tan1x+cot1x=π
2tan1x=ππ2 [Scince,tan1x+cot1x=π2]
tan12x1x2=π2 [Scince,2tan1x=tan12x1x2]
2x1x2=tanπ2
2x1x2=tan10
Cross multiplying
1x2=0
x2=±1
Here only x=1 satifies the given equation.
Note:- Here ,Putting x=-1 in the given equation we get,
3tan1(1)+cot1(1)=π
3tan1[tan(π4)]+cot1[cot(π4)]=π
3tan1[tan(π4)]+cot1[cot(π4)]=π
3tan1[tan(π4)]+πcot1[cot(π4)]=π
3×π4+ππ4=π
π+π=π
0π
Hence , x=-1 does not satisfy the given equation.

Question:23

The value of sin1[cos(33π5)] is
A.3π5
B.7π5
C.π10
D.π10

Answer:

Answer :(D)
We have
sin1[cos(33π5)]
=sin1[cos(6π+3π5)]
=sin1[cos(3π5)] [Since,cos(2nπ+θ)=cosθ]
=sin1[cos(π2+π10)]
=sin1[sin(π10)][Since,sin1(x)=sin1x]
=π10[Since,sin1(sinx)=x,xϵ(π2,π2)]

Question:24

The domain of the function cos1(2x1) is
A.[0,1]
B.[-1,1]
C.(-1,1)
D.[0,π]

Answer:

Answer:(A)
We Have f(x)=cos1(2x1)
Scince 12x11
02x2
0x1
xϵ[0,1]

Question:25

The domain of the function defined by f(x)=sin1x1 is
A.[1,2]
B.[-1,1]
C.[0,1]
D. None of these

Answer:

Answer: (A)
f(x)=sin1x1
0x11[Since,x10and1x11]
1x2
xϵ[1,2]

Question:26

If cos(sin125+cos1x)=0, then x is equal to
A.15
B.25
C.0
D.1

Answer:

Answer: (B)
Given, cos(Sin125+cos1x)=0
Let Sin125+cos1x=θ
So cosθ=0.......(1)
Principal value cos1x is [0,π].......(2)
Also , we know that cosπ2=0......(3)
From (1) ,,(2), and (3) we have
θ=π2
But θ=sin125+cos1x
So,
sin125+cos1x=π2
We know that sin1x+cos1x=π2forallxϵ[1,1]
As sin125+cos1x=π2
so x=25

Question:27

The value of sin (2tan–1 (.75)) is equal to
A. 0.75
B. 1.5
C. 0.96
D. sin 1.5

Answer:

Answer :(c)
sin (2tan–1 (.75))
Let, tan–1 (.75) = θ
tan1(34)=θ
tanθ=34
As, tanθ=34 so
sinθ=35,cosθ=45......(1)
Now,
sin (2tan–1 (.75)) = sin 2θ
= 2 sin θ cos θ
=2(35)(45)
=2425
So, sin (2tan–1 (.75)) = 0.96.

Question:28

The Value of cos1cos3π2 is equal to
A.π2
B.3π2
C.5π2
D.7π2

Answer:

We have cos1cos3π2
We know that,
cos3π2=0
So, cos1cos3π2=cos10
Let cos10=θ
⇒ cos θ = 0
Principal value of cos-1 x is [0, π]
For, cos θ = 0
so,θ=π2

Question:29

The value of the expression 2sec12+sin1(12) is
A.π6
B.5π6
C.7π6
D.1

Answer:

Answer :(B)
We have,
Principal value of sin-1 x is (π2,π2)
Principal value of sec-1 x is [0, π]{π2}
Let sin112=A
sinA=12
A=π6
So, sin112=π6 … (1)

Let sec-1 2 = B
⇒ sec B = 2
B=π3
So, 2 sec-1 2 = 2B
2sec12=2π3...(2)
So, the value of 2sec12+sin112 from (1) and (2) is
2sec12+sin112=2π3+π6
=4π6+π6
=5π6
So, 2sec12+sin112=5π6

Question:30

If tan–1 x + tan–1 y = 4π/5, then cot–1x + cot–1 y equals
A.π5
B.2π5
C.3π5
D.π

Answer:

Answer :(A)
We know that,
tan1x+cot1x=π2
We have,
tan–1 x + tan–1 y = 4π/5 … (1)
Let, cot–1x + cot–1 y = k … (2)
Adding (1) and (2) –
tan1x+tan1y+cot1x+cot1y=4π5+k...(3)
Now, tan–1 A + cot–1 A = π/2 for all real numbers.
So, (tan–1 x + cot–1 x) + (tan–1y + cot–1 y) = π … (4)
From (3) and (4), we get,
4π5+k=π
k=π4π5
k=π5

Question:31

If sin12a1+a2+cos11a21+a2=tan12x1x2 where a, xϵ[0,1] then the value of x is
A. 0
B. a/2
C. a
D. 2a1a2

Answer:

Answer:(D)
We have
sin12a1+a2+cos11a21+a2=tan12x1x2
we know that
2tan1p=sin12p1+p2.........(1)
Also,2tan1p=cos11p21+p2.........(2)
Also,2tan1p=tan12p1p2.........(3)
From (1) and (2) we have,
L.H.S-
sin12a1+a2+cos11a21+a2=2tan1a+2tan1a
sin12a1+a2+cos11a21+a2=4tan1a
From (3) R.H.S
tan12x1x2=2tan1x
So, we have 4 tan-1 a = 2 tan-1 x
⇒ 2 tan-1 a = tan-1 x
But from (3) 2tan1a=tan12a1a2
So tan12a1a2=tan1x
x=2a1a2

Question:32

The value of cotcos1725is
A. 25/24
B.25/7
C.24/25
D.7/24

Answer:

Answer :(d)
We have to find cotcos1725
Let cos1725=A
cos1=725
Also, cotA=cotcos1725
As, sinA=1cos2A
So sinA=1(75)2
sinA=149625
sinA=62549625
sinA=576625
sinA=2425
We need to find cot A
cotA=cosAsinA
cotA=7252425
cotA=724
So cotcos1725=724

Question:33

The value of the expression tan12cos125 is [Hint:tanθ2=1cosθ1+cosθ]
A.2+5
B.52
C.2+52
D.5+2

Answer:

Answer:(B)
We need to find , tan12cos125
Let, cos125=A
cosA=25
Also we need to find tanA2
We know that tanθ2=(1cosθ)1+cosθ
so, tan1A2=(1cosA)1+cosA
tan1A2=1251+25
tan1A2=5255+25
tan1A2=525+2
on rationalizing,
tan1A2=(52)(5+2)(5+2)(5+2)
tan1A2=(5)222(52+22)
tan1A2=54(52+22)
tan1A2=1(52+22)
tan1A2=15+2
Again rationalizing
tan1A2=1(51)(5+2)(52)
tan1A2=(52)(5222)
tan1A2=(52)(54)
tan1A2=52

Question:34

If |x| ≤ 1, then 2tan2x+sin12x1+x2 is equal to
A. 4 tan–1 x
B. 0
C. π2
D. π

Answer:

Answer(A)
We need to find, 2tan1x+sin12x1+x2
We know that
2tan1p=sin12x1+x2
So,
2tan1x+sin12x1+x2=2tan1x+2tan1x
=4tan1x

Question:35

If cos–1α + cos–1β + cos–1γ = 3π, then α(β + γ) + β (γ + α) + γ (α + β) equals
A. 0
B. 1
C. 6
D. 12

Answer:

Answer :(C)
Given, cos–1α + cos–1β + cos–1γ = 3π … (1)
Principal value of cos-1 x is [0, π]
So, maximum value which cos-1 x can have is π.
So, if (1) is correct then all the three terms i.e,
cos–1α, cos–1β, cos–1γ should be equal to π
So, cos–1α = π
cos–1β = π
cos–1γ = π
So, α = β = γ = -1
So, α(β + γ) + β (γ + α) + γ (α + β)
= (-1)(-1-1) + (-1)(-1-1) + (-1)(-1-1)
= 3(-1)(-2)
= 6

Question:36

The number of real solutions of equarion 1+cosx=2cos1(cosx)in[π2,π] is
A. 0
B. 1
C. 2
D. Infinite

Answer:

Answer:(A)
We have , 1+cos2x=2cos1(cosx), x is in [π2,π]
R.H.S
2cos1(cosx)=2x
So, 1+cos2x=2x
Squaring both side , we get,
(1+cos2x)=2x2
cos2x=2x21
Now plotting cos 2x and 2x2-1, we get,
a36
As , there is no point of intersection in [π2,π], so therre is no
solution of the given equation in [π2,π]

Question:37

If cos1x>sin1x , then
A.12<x1
B.0x<12
C.1x<12
D.x>0

Answer:

Answer :(C)
Plotting cos-1 x and sin-1 x, we get,
a37
As, graph of cos-1 x is above graph of sin-1 x in [1,12).
So, cos–1x > sin–1 x for all x in [1,12) .

Question:38

Fill in the blanks The principle value of cos1(12) is ___________.

Answer:

The principal value of cos1(12) is 2π3.
Principal value cos-1 x is [0,π]
Let, cos1(1)=θ
cosθ=12
As, cos2π3=12
So, θ=2π3

Question:39

Fill in the blanks The value of sin1(sin3π5) is_______.

Answer:

The value of sin1(sin3π5) is 2π5
Principal value of sin1 is [π2,π2]
now, sin1(sin3π5) should be in the given range
3π5 is outside the range [π2,π2]
As, sin (π – x) = sin x
So, sin1(sin3π5)=sin1(sin(π3π5))
=sin1(sin2π5)
=sin1(sin2π5)=2π5

Question:40

Fill in the blanks
If cos (tan–1x + cot–1 √3) = 0, then value of x is _________.

Answer:

 If cos(tan1x+cot13)=0, then value of x is 3 .  Given, cos(tan1x+cot13)=0tan1x+cot13=π2 We know that, tan1x+cot1x=π2 So, x=3

Question:40

Fill in the blanks
If cos (tan–1x + cot–1 √3) = 0, then value of x is _________.

Answer:

If cos (tan–1x + cot–1 √3) = 0, then value of x is 3
Given, cos (tan–1x + cot–13) = 0
tan1x+cot13=π2
we know that, tan1x+cot1x=π2
so, x=3

Question:41

fill in blanks the set of value of sec1(12) is______________.
Answer:

Fill in the blanks the set of value of sec1(12) is ϕ
Domain of sec-1 x is R – (-1,1).
As, 12 is outside domain of sec-1 x.
Which means there is no set of value of sec112
So, the solution set of sec112 is null set or ϕ

Question:42

Fill in the blanks
The principal value of tan–1 √3 is _________.

Answer:

The Principal value of tan13 is π3
Principal value of tan-1 x is (π2,π2)
Let, tan1(3)=θ
tanθ=3
As tanπ3=3
so, θ=3

Question:43

The value of cos1(cos14π3)

Answer:

The value of cos1(cos14π3) is 2π3
We needd, cos1(cos14π3)
Principal value of cos-1 x is [0,π]
Also, cos (2nπ + θ) = cos θ for all n ? N
cos14π3=cos(4π+2π3)
cos14π3=cos2π3
So,cos1(cos14π3)=cos1(cos2π3)
cos1(cos14π3)=2π3

Question:44

Fill in the blanks
The value of cos (sin–1 x + cos–1 x), |x| ≤ 1 is ________.

Answer:

The value of cos (sin–1 x + cos–1 x) for |x| ≤ 1 is 0.
cos (sin–1 x + cos–1 x), |x| ≤ 1
We know that, (sin–1 x + cos–1 x), |x| ≤ 1 is π2
So, cos(sin1x+cos1x)=cosπ2
= 0

Question:45

The value of expression tan(sin1x+cos1x2), when x=32 is___________.

Answer:

The value of expression tan(sin1x+cos1x2) When X=32 is 1
tan(sin1x+cos1x2) When X=32
We know that, (sin–1 x + cos–1 x) for all |x| ≤ 1 is π2
As, x=32 lies in domain
So tan(sin1x+cos1x2)=tanπ4
=1

Question:46

Fill in the blanks if y=2tan1x+sin12x1+x2 for all x, then ______<y<_____.

Answer:

Fill in the blanks if y=2tan1x+sin12x1+x2 for all x, then 2π<y<2π
y=2tan1x+sin12x1+x2
We know that,
2tan1p=sin12x1+x2
so
2tan1x+sin12x1+x2=2tan1x+2tan1x
=4 tan-1 x
So, y = 4 tan-1 x
As, principal value of tan-1 x is (π2,π2)
So, 4tan1xϵ(2π,2π)
Hence, -2π < y < 2π

Question:47

The result tan1xtan1(xy1+xy) is true when value of xy is _________.

Answer:

The result tan1xtan1(xy1+xy) is true when value of xy is > -1.
We have,
tan1xtan1=tan1xy1+xy
Principal range of tan-1a is (π2,π2)
Let tan-1x = A and tan-1y = B … (1)
So, A,B ϵ(π2,π2)
We know that, tan(AB)=tanAtanB1tanAtanB … (2)
From (1) and (2), we get,
Applying, tan-1 both sides, we get,
tan1tan(AB)=tan1xy1xy
As, principal range of tan-1a is (π2,π2)
So, for tan-1tan(A-B) to be equal to A-B,
A-B must lie in (π2,π2)– (3)
Now, if both A,B < 0, then A, B ϵ(π2,0)
∴ A ϵ(π2,0) and -B ϵ(0,π2)
So, A – B ϵ(π2,π2)
So, from (3),
tan-1tan(A-B) = A-B
tan1xtan1y=tan1xyz+xy
Now, if both A,B > 0, then A, B ϵ(0,π2)
∴ A ϵ(0,π2) and -B ϵ(π2,0)
So, A – B ϵ(π2,π2)
So, from (3),
tan-1tan(A-B) = A-B
tan1xtan1y=tan1xyz+xy
Now, if A > 0 and B < 0,
Then, A ϵ(0,π2) and B ϵ(0,π2)
∴ A ϵ(0,π2) and -B ϵ(0,π2)
So, A – B ϵ (0,π)
But, required condition is A – B ϵ (π2,π2)
As, here A – B ϵ (0,π), so we must have A – B ϵ (0,π2)
AB<π2
A<π2+B
Applying tan on both sides,
tanA<tan(π2+B)
As, tan(π2+α)=cotα
So, tan A < - cot B
Again, cotα=1tanα
So, tanA<1tanB
⇒ tan A tan B < -1
As, tan B < 0
xy > -1
Now, if A < 0 and B > 0,
Then, A ϵ (π2,0) and B ϵ (0,π2)
∴ A ϵ (π2,0) and -B ϵ (π2,0)
So, A – B ϵ (-π,0)
But, required condition is A – B ϵ (π2,π2)
As, here A – B ϵ (0,π), so we must have A – B ϵ (π2,0)
AB>π2
A>Bπ2
Applying tan on both sides,
tanA>tan(Bπ2)
As, tan(απ2)=cotα
So, tan B > - cot A
Again, cotα1tanα
So, tanB>1tanA
⇒ tan A tan B > -1
⇒xy > -1

Question:48

Fill in the blanks
The value of cot–1(–x) for all x ? R in terms of cot–1 x is _______.

Answer:

The value of cot–1(–x) for all x ? R in terms of cot–1 x is
π – cot-1 x.
Let cot–1(–x) = A
⇒ cot A = -x
⇒ -cot A = x
⇒ cot (π – A) = x
⇒ (π – A) = cot-1 x
⇒ A = π – cot-1 x
So, cot–1(–x) = π – cot-1 x

Question:49

State True or False for the statement
All trigonometric functions have inverse over their respective domains.

Answer:

True.
It is well known that all trigonometric functions have inverse over their respective domains.

Question:50

State True or False for the statement
The value of the expression (cos–1x)2 is equal to sec2 x.

Answer:

As, cos-1 x is not equal to sec x. So, (cos–1x)2 is not equal to sec2 x.

Question:51

State True or False for the statement
The domain of trigonometric functions can be restricted to any one of their branch (not necessarily principal value) in order to obtain their inverse functions.

Answer:

As, all trigonometric and their corresponding inverse functions are periodic so, we can obtain the inverse of a trigonometric ratio in any branch in which it is one-one and onto.

Question:52

State True or False for the statement
The least numerical value, either positive or negative of angle θ is called the principal value of the inverse trigonometric function.

Answer:

True
We know that the smallest value, either positive or negative of angle θ is called principal value of the inverse trigonometric function

Question:53

State True or False for the statement
The graph of an inverse trigonometric function can be obtained from the graph of their corresponding trigonometric function by interchanging x and y axes.

Answer:

True.
Graph of any inverse function can be obtained by interchanging x and y-axis in the graph of the corresponding function. If (p, q) are two points on f(x) then (q, p) will be on f-1(x).

Question:54

State True or False for the statement
The minimum value of n for which tan1nπ>π4,nϵN is valid is 5.

Answer:

false
tan1nπ>π4
As , tan is an increasing function so applying tan on both side
we get,
tan(tan1nπ)>tanπ4
As, tan(tan1nπ)=nπ and tanπ4=1
so nπ>1
⇒ n > π
⇒ n > 3.14
As, n is a natural number, so least value of n is 4.

Question:55

State True or False for the statement

The principal value of sin1[cos(sin112)] is π3

Answer:

True
Principal value of sin-1 x is [π2,π2]
Principal value of cos-1 x is [0, π]
We have, sin1[cos[sin1(12)]]
As, sinπ6=12 so
sin1[cos[sin1(12)]]=sin1[cos[sin1(sinπ6)]]
sin1[cos[sin1(12)]]=sin1[cos[π6]]
As, cosπ6=32 so,
sin1[cos[sin1(12)]]=sin1[sin[π3]]
sin1[cos[sin1(12)]]=π3

Sub-Topics Covered in NCERT Exemplar Class 12 Maths Solutions Chapter 2

The sub-topics that are covered in this chapter of inverse trigonometric functions are:

  • Introduction
  • Basic concepts
  • Properties of inverse trigonometric functions
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NCERT Exemplar Class 12 Maths Solutions

Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 2

Class 12 Math NCERT Exemplar solutions chapter 2, students will get detailed answers to the questions in the NCERT book after every topic. Understanding and grasping this chapter can help one aim for a better score in their school exams, boards and their entrance exams.

  • In NCERT Exemplar solutions for Class 12 Math chapter 2, cover properties and graphical representations of inverse trigonometric functions.
  • One will learn about the necessity of studying inverse trigonometric functions and their properties. It covers the basic details about inverse trigonometric functions.
  • These solutions provide plenty of questions to practice.
JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

NCERT solutions of class 12 - Subject-wise

Here are the subject-wise links for the NCERT solutions of class 12:

NCERT Notes of class 12 - Subject Wise

Given below are the subject-wise NCERT Notes of class 12 :

NCERT Books and NCERT Syllabus

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NCERT Exemplar Class 12 Solutions - Subject Wise

Given below are the subject-wise exemplar solutions of class 12 NCERT:


Frequently Asked Questions (FAQs)

1. What are the important topics of this chapter?

Introduction to Inverse Trigonometric Functions, The Basic Concepts of Inverse Trigonometric Functions and Properties of Inverse Trigonometric Functions are important topics of this chapter.

2. Are these solutions helpful for board examinations?

Yes, the NCERT exemplar Class 12 Maths chapter 2 solutions are helpful for you to prepare for board exams.

3. How many questions are there in this chapter?

There is only 1 exercise in this chapter with 55 problem solving questions.

4. Are these solutions helpful for competitive examinations?

Yes, NCERT exemplar solutions for Class 12 Maths chapter 2 cover syllabus for exams are very reliable for preparing for competitive entrance exams like NEET and JEE Main.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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