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NCERT Exemplar for Class 12 Maths Chapter-2 (Book Solutions)

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NCERT Exemplar for Class 12 Maths - Inverse Trigonometric Functions - Free PDF Download

Free PDF download of NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions solved by expert Maths teachers on Vedantu.com as per NCERT (CBSE) Book guidelines. All Chapter 2 - Inverse Trigonometric Functions Exercise questions with solutions to help you to revise complete syllabus and score more marks in your Examinations.

Students will gain a full understanding of the subjects covered in their Class by referring to the NCERT Exemplar textbooks for various Classes subject-wise. Additionally, students can use NCERT Exemplar Solutions to discover how to do the Exercise questions in each Chapter. All of these answers are developed by subject matter specialists using the most recent CBSE patterns.

In NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions)  Students will learn about the properties and principles of Inverse Trigonometry Functions in this Chapter. They'll also learn about the limitations of Trigonometric Functions' ranges and domains. They will also use pictorial representations to document their actions. Students who want to understand how to solve these Exercise questions correctly can get the NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions) solutions PDF from the link on the Vedantus site.

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Competitive Exams after 12th Science

Access NCERT Exemplar Solutions for Class 11 Mathematics Chapter 2 - Inverse Trigonometric Functions

EXAMPLES:

Solved Examples Short Answer Questions:

1: Find the principal value of cos1x, for x=32.

Ans:  Given: x=32

The value of the inverse trigonometric function which lies in the range of the principal branch is its principal value.

Assume cos1(32)=θ, then cosθ=32.

Considering principal branch, θ[0,π]

32>0,θ being in the first quadrant

Hence,cos1(32)=π6.


Example 2: Evaluate tan1(sin(π2)).

Ans:  Given: tan1(sin(π2)).

Using the property of trigonometric inverse, sin1(x)=sin1x

tan1(sin(π2))

tan1(sin(π2))

tan1(1)=π4 


Example 3: Find the value of cos1(cos13π6).

Ans:  Given: cos1(cos13π6)

The value of the inverse trigonometric function which lies in the range of principal branch is its principal value.

cos1(cos13π6)

cos1(cos(2π+π6))(cos(2π+θ) = cosθ)

cos1(cosπ6)=π6(cos1(cos θ )=θ) 

=π6.


Example 4: Find the value of tan1(tan9π8).

Ans: Given: tan1(tan9π8)

The value of the inverse trigonometric function which lies in the range of principal branch is its principal value.

tan1(tan9π8)

tan1tan(π+π8)(tan(π+θ)=tanθ)

tan1(tan(π8))=π8(tan1(tanθ)=θ) 


Example 5: Evaluate tan(tan1(4))

Ans:  Given: tan(tan1(4))

 Using the property of trigonometric inverse, tan1(x)=tan1x

  tan(tan1x)=x,xR,

  tan(tan1(4))=4 


Example 6: Evaluate: tan13sec1(2).

Ans:  Given: tan13sec1(2)

Using the property of trigonometric inverse, sec1(x)=πsec1x


Example 7: Evaluate: sin1[cos(sin132)].

Ans:  Given sin1[cos(sin132)]

The value of the inverse trigonometric function which lies in the range of the principal branch is its principal value.

sin1[cos(sin132)]

sin1[cos(π3)]

sin1[12]=π6 


Example 8: Prove that tan(cot1x)=cot(tan1x). State with reason whether the equality is valid for all values of x

Ans:  Given tan(cot1x)=cot(tan1x)

Using the property of trigonometric inverse,  tan(tan1x)=x

Assume cot1x=θ. cotθ=x

tan(π2θ)=x

tan1x=π2θ 

So,

  tan(cot1x)=tanθ

cot(π2θ)

cot(π2cot1x)

   =cot(tan1x) 

The equality is valid for all values of x since tan1x and cot1x is true forxR.


Example 9: Find the value of sec(tan1y2)

Ans:  Given: sec(tan1y2)

Assumetan1y2=θ, whereθ(π2,π2)

So, tanθ=y2

secθ=4+y22

sec(tan1y2)=secθ

=4+y22 


Example 10: Find value of tan(cos1x) and hence evaluate tan(cos1817).

Ans:  Given: tan(cos1x),tan(cos1817)

Using the property of trigonometric inverse,  cos(cos1x)=x

Assumecos1x=θ, cosθ=x, where θ[0,π]

tanθ=1x2x

tan(cos1817) 

tan(cos1x)

tanθ=1cos2θcosθ

=1(817)2817=158


Example 11: Find the value of sin[2cot1(512)]

Ans:  Given: sin[2cot1(512)]

Using the property of trigonometric inverse.

Assumecot1(512)=y. Then coty=512

Now,

sin[2cot1(512)]=sin2y

Since,coty<0, so [y(π2,π)]

2sinycosy=2(1213)(513)

=120169 


Example 12: Evaluate cos[sin114+sec143]

Ans:  Given: cos[sin114+sec143]

Using the property of trigonometric inverse sin1x+cos1x=π2:x[1,1]

cos[sin114+sec143]

cos[sin114+cos134]

cos(sin114)cos(cos134)sin(sin114)sin(cos134)

341(14)2141(34)2

341541474=315716 


Example 13: Prove that 2sin135tan11731=π4

Ans:  Given: 2sin135tan11731=π4

Using the property of trigonometric inverse, y=sin1x,x[π2,π2]

Assumesin135=θ, then sinθ=35, where θ[π2,π2]

  tanθ=34

θ=tan134 

Therefore,

2sin135tan11731

2θtan11731

2tan134tan11731

tan1(2341916)tan11731

tan1247tan11731

tan1(24717311+2471731)=π4 


Example 14: Prove that cot17+cot18+cot118=cot13

Ans:  Given: cot17+cot18+cot118=cot13

Using the property of trigonometric inverse cot1x=tan11x, if x>0

cot17+cot18+cot11

tan117+tan118+tan1118

Since, xy=1718<1

=tan1(17+18117×18)+tan1118

Since xy<1

tan1311+tan1118

tan1(311+1181311×118)

tan165195

tan113=cot13 


Example 15: Which is greater, tan 1 or tan11 ?


seo images


Ans:  Given:  figure

The value of the inverse trigonometric function which lies in the range of the principal branch is its principal value.

From Fig., tan x is an increasing function in the interval(π2,π2),

Since,1>π4tan1>tanπ4

tan1>1

tan1>1>π4

tan1>1>tan1(1)


Example 16: Find the value sin(2tan123)+cos(tan13)

Ans:  Given: sin(2tan123)+cos(tan13)

Using the trigonometric inverse function, y=tan1x,x[π2,π2]

Assume tan123=x and tan13=y

tanx=23 And tany=3

Therefore,

sin(2tan123)+cos(tan13)

sin(2x)+cosy

2tanx1+tan2x+11+tan2y

2231+49+11+(3)2

1213+12=3726 


Example 17: Solve for xtan1(1x1+x)=12tan1x,x>0

Ans:  Given: tan1(1x1+x)=12tan1x,x>0

Using the trigonometric inverse function, y=tan1x,x[π2,π2]

  tan1(1x1+x)=12tan1x

2tan1(1x1+x)=tan1x 

2[tan11tan1x]=tan1x

2(π4)=3tan1x 

π6=tan1x

x=13 


Example 18: Find the values of x which satisfy the equation sin1x+sin1(1x)=cos1x.

Ans:  Given: sin1x+sin1(1x)=cos1x 

Using the trigonometric inverse function, y=sin1x,x[π2,π2]

Taking the sin on both sides of sin1x+sin1(1x)=cos1x 

sin(sin1x+sin1(1x))=sin(cos1x)

sin(sin1x)cos(sin1(1x))+cos(sin1x)sin(sin1(1x))=sin(cos1x)

x1(1x)2+(1x)1x2=1x2

x2xx2+1x2(1x1)=0

x(2xx21x2)=0

x=0or2xx2=1x2

x=0orx=12.


Example 19: Solve the equation sin16x+sin163x=π2

Ans:  Given: sin16x+sin163x=π2

Using the trigonometric inverse function, y=sin1x,x[π2,π2]

 sin16x=π2sin163x

sin(sin16x)=sin(π2sin163x)

6x=cos(sin163x)

6x=1108x2.

Squaring, 

36x2=1108x2

144x2=1

x=±112 

Here, x=112 is the only root of the equation as x=112 does not satisfy it.


Example 20: Show that 2tan1{tanα2×tan(π4β2)}=tan1sinαcosβcosα+sinβ

Ans:  Given: 2tan1{tanα2tan(π4β2)}=tan1sinαcosβcosα+sinβ

Using the property of trigonometric inverse (2tan1x=tan12x1x2)

Taking L.H.S. =tan12tanα2tan(π4β2)1tan2α2tan2(π4β2)

=tan12tanα2(1tan2β2)(1+tanβ2)2tan2α2(1tanβ2)2

=tan12tanα2(1tan2β2)(1+tan2β2)(1tan2α2)+2tanβ2(1+tan2α2)

=tan12tanα21+tan2α21tan2β21+tan2β21tan2α21+tan2α2+2tanβ21+tan2β2

=tan1(sinαcosβcosα+sinβ)

=R.H.S. 

L.H.S=R.H.S 

Hence, proved.


Objective Type Questions Choose the Correct Answer from the Given Four Options in Each of the Examples 21 to 41

Example 21: Which of the following corresponds to the principal value branch of tan1?

(A) (π2,π2)

(B) [π2,π2]

(C) (π2,π2){0}

(D) (0,π)

Ans: Correct answer is option A.

Given: tan1

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

The principal value branch of tan1 is (π2,π2).

Hence, (π2,π2) is the correct answer.


22. The principal value branch of sec1 is

(A) [π2,π2]{0}

(B) [0,π]{π2}

(C) (0,π)

(D) (π2,π2)

Ans:  Correct answer is option B.

Given: sec1

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

The principal value branch of sec1 is [0,π]{π2}.

Hence, [0,π]{π2} is the correct answer.


23. One branch of cos1 other than the principal value branch corresponds to

(A) [π2,3π2]

(B) [π,2π]{3π2}

(C) (0,π)

(D) [2π,3π]

Ans: Correct answer is option D.

Given: cos1

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

One branch of cos1 other than the principal value branch corresponds to is [2π,3π].

Hence, [2π,3π] is the correct answer.


24. The value of sin1(cos(43π5)) is

(A) 3π5

(B) 7π5

(C) π10

(D) π10

Ans:  Correct answer is option D.

Given: sin1(cos(43π5))

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

sin1(cos(43π5))=sin1(cos(40π+3π5))

sin1(cos(43π5))=sin1(cos(8π+3π5))

As (8π+θ) is in the first quadrant, coswill be positive.

sin1(cos(43π5))=sin1(cos(3π5))

sin1(cos(43π5))=sin1(sin(π23π5))

sin1(cos(43π5))=sin1(sin(π10))

sin1(cos(43π5))=π10

Hence, sin1(cos(43π5))=π10.


25. The principal value of the expression cos1[cos(680o)] is

(A) 2π9

(B) 2π9

(C) 34π9

(D) π9

Ans: Correct answer is option A.

Given: cos1[cos(680)]

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

cos1[cos(680)]=cos1[cos(72040)]

cos1[cos(680)]=cos1[cos(40)]

As cos(θ)=cosθ,

cos1[cos(680)]=cos1[cos(40)]

cos1[cos(680)]=40

cos1[cos(680)]=2π9

Hence, cos1[cos(680)]=2π9.


26. The value of cot(sin1x) is

(A) 1+x2x

(B) x1+x2

(C) 1x

(D) 1x2x

Ans:  Correct answer is option D.

Given: cot(sin1x)

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

Let sin1x=θ

sinθ=x

cosecθ=1x

cosec2θ=1x2

As cosec2θcot2θ=1,

1+cot2θ=1x2

cot2θ=1x21

cot2θ=1x2x2

cotθ=1x2x2

cotθ=1x2x

cot(sin1x)=1x2x

Hence, cot(sin1x)=1x2x.


27. If tan1x=π10 for some xR, then the value of cot1x is

(A) π5

(B) 2π5

(C) 3π5

(D) 4π5

Ans:  Correct answer is option B.

Given: tan1x=π10 for some xR.

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

As tan1x+cot1x=π2,

cot1x=π2π10

cot1x=2π5

Hence, cot1x=2π5.


28. The domain ofsin12x is

(A) [0,1]

(B) [1,1]

(C) [12,12]

(D) [2,2]

Ans: Correct answer is option C.

Given: sin12x

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

Let sin12x=θ

2x=sinθ

As 1sinθ1,

12x1

12x12

Hence, [12,12] is the domain of sin12x.


29. The principal value of sin1(32) is

(A) 2π3

(B) π3

(C) 4π3

(D) 5π3

Ans: Correct answer is option B.

Given: sin1(32)

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

sin1(32)=sin1(sinπ3)

sin1(32)=sin1(sinπ3)

sin1(32)=π3

Hence, sin1(32)=π3.


30. The greatest and least values of (sin1x)2+(cos1x)2 are respectively

(A) 5π24 and π28

(B) π2 and π2

(C) π24 and π24

(D) π24 and 0

Ans:  Correct answer is option A.

Given: (sin1x)2+(cos1x)2

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

As a2+b2=(a+b)22ab,

(sin1x)2+(cos1x)2=(sin1x+cos1x)22sin1xcos1x

(sin1x)2+(cos1x)2=π422sin1x(π2sin1x)

(sin1x)2+(cos1x)2=π42πsin1x+2(sin1x)2

(sin1x)2+(cos1x)2=2[(sin1x)2π2sin1x+π28]

(sin1x)2+(cos1x)2=2[(sin1xπ4)2+π216]

The least value will be 2(π216)=π28.

The greatest value will be 2[(π2π4)2+π216]=5π24

Hence, the greatest and least values of (sin1x)2+(cos1x)2 are 5π24 and π28.


31. Let θ=sin1(sin(600o)), then value of θ is

(A) π3

(B) π2

(C) 2π3

(D) 2π3

Ans:  Correct answer is option A.

Given: θ=sin1(sin(600))

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

sin1(sin(600))=sin1(sin(600×π180))

sin1(sin(600))=sin1(sin(10π3))

sin1(sin(600))=sin1(sin(4π2π3))

sin1(sin(600))=sin1(sin2π3)

sin1(sin(600))=sin1(sin(ππ3))

sin1(sin(600))=sin1(sinπ3)

sin1(sin(600))=π3

Hence, sin1(sin(600))=π3..


32. The domain of the function y=sin1(x2) is

(A) [0,1]

(B) (0,1)

(C) [1,1]

(D) f

Ans: Correct answer is option C.

Given: :y=sin1(x2)

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

y=sin1(x2)

siny=x2

As 1siny1,

1x21

1x21

0x21

|x|1

1x1

Hence, the domain of y=sin1(x2) is [1,1].


33. The domain of y=cos1(x24) is

(A) [3,5]

(B) [0,π]

(C) [5,3]C[5,3]

(D) [5,3][3,5]

Ans:  Correct answer is option D.

Given: y=cos1(x24)

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

y=cos1(x24)

cosy=x24

As 1cosy1,

1x241

3x25

3|x|5

x[5,3][3,5]

Hence, the domain of y=cos1(x24) is [5,3][3,5].


34. The domain of the function defined by f(x)=sin1x+cosx is

(A) [1,1]

(B) [1,π+1]

(C) (, )

(D) f

Ans: Correct answer is option A.

Given: f(x)=sin1x+cosx

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

f(x)=sin1x+cosx

The domain of cos is R.

The domain of sin1 is [1,1].

So, the domain of sin1x+cosx will be R[1,1] i.e. [1,1].

Hence, the domain of f(x)=sin1x+cosx is [1,1].


35. The value of sin(2sin1(0.6)) is

(A) 0.48

(B) 0.96

(C) 1.2

(D) sin1.2

Ans:  Correct answer is option B.

Given: sin(2sin1(0.6))

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

Let sin1(0.6)=θ

sinθ=0.6

As cos2θ+sin2θ=1,

cos2θ+(0.6)2=1

cos2θ+0.36=1

cos2θ=0.64

cosθ=0.8

As sin2θ=2sinθcosθ,

sin(2sin1(0.6))=2(0.6)(0.8)

sin(2sin1(0.6))=0.96

Hence, sin(2sin1(0.6))=0.96.


36. If sin1x+sin1y=π2, then value of cos1x+cos1y is

(A) π2

(B) π

(C) 0

(D) 2π3

Ans:  Correct answer is option A

Given: sin1x+sin1y=π2

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

Given that sin1x+sin1y=π2

(π2cos1x)+(π2cos1y)=π2

cos1x+cos1y=π2

Hence, cos1x+cos1y=π2.


37. The value of tan(cos135+tan114) is

(A) 198

(B) 819

(C) 1912

(D) 34

Ans: Correct answer is option B

Given: tan(cos135+tan114)

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

tan(cos135+tan114)=tan(tan143+tan114)

As tan1x+tan1y=tan1(x+y1xy),

tan(cos135+tan114)=tantan1(43+14143×14)

tan(cos135+tan114)=tantan1(198)

tan(cos135+tan114)=198

Hence, tan(cos135+tan114)=198.


38. The value of the expression sin[cot1(cos(tan11))] is

(A) 0

(B) 1

(C) 13

(D) 23

Ans:  Correct answer is option D

Given: sin[cot1(cos(tan11))]

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

sin[cot1(cos(tan11))]=sin[cot1(cosπ4)]

sin[cot1(cos(tan11))]=sin[cot112]

sin[cot1(cos(tan11))]=sin[sin123]

sin[cot1(cos(tan11))]=23

Hence, sin[cot1(cos(tan11))]=23.


39. The equation tan1xcot1x=tan1(13) has

(A) no solution

(B) unique solution

(C) infinite number of solutions

(D) two solutions

Ans:  Correct answer is option B

Given: tan1xcot1x=tan1(13)

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

tan1xcot1x=tan1(13)

tan1xcot1x=π6 ...(1)

And tan1x+cot1x=π2 ...(2)

Add (1) and (2),

2tan1x=2π3

tan1x=π3

x=3

Hence, tan1xcot1x=tan1(13) has a unique solution.


40. If α2sin1x+cos1xβ, then

(A) α=π2,β=π2

(B) α=0,β=π

(C) α=π2,β=3π2

(D) α=0,β=2π

Ans:  Correct answer is option B

Given: α2sin1x+cos1xβ

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

As π2sin1xπ2,

π2+π2sin1x+π2π2+π2

0sin1x+(sin1x+cos1x)π

02sin1x+cos1xπ

Hence, α=0,β=π.


41. The value of tan2(sec12)+cot2(cosec13) is

(A) 5

(B) 11

(C) 13

(D) 15

Ans:  Correct answer is option B

Given: tan2(sec12)+cot2(cosec13)

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

As cosec2θcot2θ=1 and sec2θtan2θ=1,

tan2(sec12)+cot2(cosec13)=sec2(sec12)1+cosec2(cosec13)1

tan2(sec12)+cot2(cosec13)=22×1+322

tan2(sec12)+cot2(cosec13)=11

Hence, tan2(sec12)+cot2(cosec13)=11.


EXERCISE:

Short Answer Questions:

1. Find the value of tan1(tan5π6)+cos1(cos13π6).

Ans: Given: tan1(tan5π6)+cos1(cos13π6)

Using the property of trigonometric inverse, tan1tanx=x,x(π2,π2)

from the relation, tan1tanx=x,x(π2,π2)And,

tan1(tan5π6)5π65π6(π2,π2)

As, cos1cosx=x;x[0,π]

cos1(cos13π6)13π613π6[0,π]

Solving the given expression,tan1(tan5π6)+cos1(cos13π6)

=tan1[tan(ππ6)]+cos1[cos(2π+π6)]

=tan1(tanπ6)+cos1(cos7π6)

=tan1(tan(π6))+[cos1cos(π6)]

=π6+π6=0 


2. Evaluate cos[cos1(32)+π6].

Ans:  Given: cos[cos1(32)+π6].

Using the property of trigonometric inverse, cos1cosx=x;x[0,π]

On solving for given expression,

cos[cos1(32)+π6]

=cos[cos1(cos5π6)+π6](cos5π6=32)

=cos(5π6+π6)(cos1cosx=x;x[0,π])

=cos(π)=1


3. Prove that cot(π42cot13)=7.

Ans:  Given: cot(π42cot13)=7

Using the property of trigonometric inverse, 2tan1x=tan1(2x1x2)

Solving the given relation cot(π42cot13)=7

π42cot13=cot17

2tan113=π4tan117

2tan113+tan117=π4

Now, solving 2tan113+tan117

=tan1231(13)2+tan117(2tan1x=tan12x1x2)

=tan134+tan117

=tan134+1713417

=tan1(21+4)28(28328)

=tan12525

=tan11=π4 


4. Find the value of tan1(13)+cot1(13)+tan1(sin(π2))

Ans:  Given: tan1(13)+cot1(13)+tan1(sin(π2))

Using the property of trigonometric inverse,tan(tan1x)=x,tan1(x)=tan1x,xR

tan1(13)+cot1(13)+tan1(sin(π2))

tan1(tan(π6))+cot1(cotπ3)+tan1(1)

π6+π3+(π4)=π12 


5. Find the value of tan1(tan2π3).

Ans:  Given: tan1(tan2π3)

Using the property of trigonometric inverse, tan1tanx=x,x(π2,π2)

=tan1(tan2π3)

=tan1tan(ππ3)

=tan1(tan(π3))=π3 


6. Show that 2tan1(3)=π2+tan1(43)

Ans:  Given:  2tan1(3)=π2+tan1(43)

Using the property of trigonometric inverse,  tan(tan1x)=x,tan1(x)=tan1x,xR

Solving for L.H.S,

L.H.S,

2tan1(3)=2tan13(tan1(x)=tan1x,xR)

=2[π2cot13](tan1x+cot1x=π2)

=2[π2tan113](tan1x=cot11x,x>0)

=π+2tan113 

Applying 2tan1x=tan12x1x2

=π+tan12131(13)2

=π+tan12/38/9=π+tan134

=π+π2cot134(tan1x+cot1x=π2)

=π2tan143(tan1x=cot11x,x>0)

=π2+tan1(43)(tan1(x)=tan1x,xR)


7. Find the real solutions of the equation tan1x(x+1)+sin1x2+x+1=π2

Ans: Using the property of trigonometric inverse,  tan(tan1x)=x

tan1x(x+1)+sin1x2+x+1=π2

Solving 

tan1x(x+1)=π2sin1x2+x+1

tan1x2+x=cos1x2+x+1[sin1x+cos1x=π2]

cos1[11+x2+x]=cos1x2+x+1[tan1x=cos111+x2]

1x2+x+1=x2+x+1

x2+x+1=1

x2+x=0 

x(x+1)=0

x=0 

Also,x+1=0

x=0,x=1


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

Ans:  Given: sin(2tan113)+cos(tan122)

Using the property of trigonometric inverse, sin(sin1x)=x,x[1,1]

Solving for L.H.S and R.H.S separately,

L.H.S = sin(2tan113)

=sin(sin12×131+(13)2)(2tan1x=sin12x1+x2) 

=sin(sin123109)

=sin(sin135)=35(sin(sin1x)=x,x[1,1]) 

R.HS.=cos(tan122)

cos(cos113)=13(cos(cos1x)=x,x[1,1])

 Adding the values of L.H.S and R.H.S

sin(2tan113)+cos(tan122)

=35+13

=9+515=1415 


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

Ans:  Given: 2tan1(cosθ)=tan1(2cosecθ)

Using the property of trigonometric inverse, 2tan1x=tan1(2x1x2)

tan1(2cosθ1cos2θ)=tan1(2cosecθ)(2tan1x=tan1(2x1x2))

2cosθsin2θ=2cosecθ

2cosθsin2θ=2sinθ

cosθsinθ=1

cotθ=1

θ=π4 


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

Ans:  Given: cos(2tan117)=sin(4tan113)

Using the property of trigonometric inverse, 2tan1x=cos11x21+x2, cos(cos1x)=x,x[1,1], sin(sin1x)=x,x[1,1]

Solving for L.H.S and R.H.S separately,

L.H.S.=cos(2tan117)

=cos(cos11(17)21+(17)2)(2tan1x=cos11x21+x2)

=cos(cos148495049)

=cos(cos12425)=2425(cos(cos1x)=x,x[1,1]) 

R.H.S

=sin(2(2tan113))

=sin(2(tan12131(13)2))(2tan1xtan1(2x1x2))

=sin(2tan12389)

=sin(2tan134) 

=sin(sin12×341+(34)2)(2tan1x=sin12x1+x2)

=sin(sin1322516)

=sin(sin12425)=2425(sin(sin1x)=x,x[1,1]) 

L.H.S.  = R.H.S.


11. Solve the equation cos(tan1x)=sin(cot134).

Ans:  Given: cos(tan1x)=sin(cot134)

Using the property of trigonometric inverse, y=sin1x,x[π2,π2]

Solving for L.H.S and R.H.S separately,

L.H.S.=cos(tan1x)

=cos(cos11x2+1)

=1x2+1 

R.H.S.=sin(cot134)

=sin(sin145)=45 

1x2+1=45

16(x2+1)=25

16x2=9 

x2=916

x=±34 


12. Prove that tan11+x2+1x21+x21x2=π4+12cos1x2.

Ans:  Given: tan11+x2+1x21+x21x2=π4+12cos1x2

Using the property of trigonometric inverse tan(tan1x)=x

Assume

x2=cos2θ

θ=12cos1x2 

Now solving for L.H.S

L.H.S. =tan1[1+cos2θ+1cos2θ1+cos2θ1cos2θ]

tan1[2cos2θ+2sin2θ2cos2θ2sin2θ]

tan1[2cosθ+2sinθ2cosθ2sinθ]

tan1[1+tanθ1tanθ]

tan1[tanπ4+tanθ1tanπ4tanθ]

tan1(tan(π4+θ))=π4+θ 

Substituting the value of θ=12cos1x2

=π4+12cos1x2

Hence, proved.


13. Find the simplified form of cos1[35cosx+45sinx],x[3π4,π4]

Ans:  Given: cos1[35cosx+45sinx],x[3π4,π4]

Using the property of trigonometric inverse, y=sin1x,x[π2,π2]

cos1[35cosx+45sinx],x[3π4,π4]

Assume cosα=35sinα=45tanα=43

cos1[35cosx+45sinx]=cos1(cosαcosx+sinαsinx]

cos1[cos(αx)]=αx=tan143x 


14.  Prove that sin1817+sin135=sin17785.

Ans:  Given: sin1817+sin135=sin17785

Using the property of trigonometric inverse, y=sin1x,x[π2,π2]

sin1817+sin135=sin17785

Solving for L.H.S

LHS=sin1817+sin135[tan1xtan1y=tan1(xy1+xy)]

=tan1815+tan134

=tan1815+341815×34 

=tan132+4560602460

=tan17736

=sin1775929+1296

=sin17785

= R.H.S. 


15. Show that sin1513+cos135=tan16316

Ans:  Given: sin1513+cos135=tan16316

Using the property of trigonometric inverse tan(tan1x)=x

Here,

sin1513=tan1512

cos135=tan143

Now, solving for L.H.S

L.H.S.=sin1513+cos135

=tan1512+tan143

=tan1512+43151243[tan1x+tan1y=tan1(x+y1xy)]

=tan115+4836362036=tan16316

=R.H.S 


16. Prove that tan114+tan129=sin115.

Ans:  Given: tan114+tan129=sin115

Using the property of trigonometric inverse tan(tan1x)=x

Solving for L.H.S

LH.S.=tan114+tan129

=tan114+2911429[tan1x+tan1y=tan1(x+y1xy)]

=tan19+8362

=tan112=sin115=R.H.S 

Hence, proved.


17. Find the value of 4tan115tan11239.

Ans:  Given: 4tan115tan11239

Using the property of trigonometric inverse tan(tan1x)=x and tan1xtan1y=tan1(xy1+xy), 2tan1x=tan12x1x2

4tan115tan11239

=2(2tan115)tan11239

=2tan1251(15)2tan11239(2tan1x=tan12x1x2) 

=2tan1252425tan11239

=tan125121(512)2tan11239

=tan1144×5119×6tan11239

=tan1120119tan11239

=tan112011912391+1201191239(tan1xtan1y=tan1xy1+xy)

=tan1120×239119119×239+120=tan12868011928441+120

=tan12856128561=tan11=π4 


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

Ans:  Given: tan(12sin134)=473

Using the trigonometric inverse function, y=sin1x,x[π2,π2]

Assume,

12sin134=θ

sin134=2θ 

sin2θ=34

Now, solving for L.H.S

2tanθ1+tan2θ=34

3tan2θ8 and θ+3=0

tanθ=8±64366

tanθ=8±286=8±276=4±73

Now π2sin134π2

π412sin134π2

tan(π4)tan(12(sin134))tanπ4

1tan(12sin134)1

tanθ=473

L.H.S = R.H.S

tanθ=4+73>1which is not possible 


19. If a1,a2,a3,,an is an arithmetic progression with common differenced, then evaluate the following expression.

tan[tan1(d1+a1a2)+tan1(d1+a2a3)+tan1(d1+a3a4)++tan1(d1+an1an)]

Ans:  Given: a1,a2,a3,,an is an arithmetic progression with common differenced

Using the property of trigonometric inverse tan(tan1x)=x and tan1xtan1y=tan1(xy1+xy)

Common difference d is, d=a2a1=a3a2=a4a3==anan1

tan1d1+a1a2=tan1a2a11+a1a2=tan1a2tan1a1

Similarly tan1d1+a2a3=tan1a3a21+a2a3=tan1a3tan1a2.....tan1d1+an1an=tan1anan11+an1an=tan1antan1an1

tan[tan1(d1+a1a2)+tan1(d1+a2a3)+.....+tan1(d1+a3a4)++tan1(d1+an1an)]

=tan[(tan1a2tan1a1)+(tan1a3tan1a2)++(tan1antan1an1)]

=tan[tan1antan1a1]

tan1xtan1y=tan1(xy1+xy)

tan[tan1ana11+ana1]

ana11+ana1 


Choose the correct answers from the given four options in each of the exercises from 20 to 37 (M.C.Q):

20. Which of the following is the principal value branch of cos1x?

(A) [π2,π2]

(B) (0,π)

(C) [0,π]

(D) (0,π){π2}

Ans: Correct answer is option C.

Given: cos1x

Use the properties of inverse trigonometric functions.

The principal value branch of cos1x is [0,π].

Therefore, [0,π] is the correct answer.


21. Which of the following is the principal value branch of cosec1x?

 (A) (π2,π2)

(B) [0,π]{π2}

(C) [π2,π2]

(D) [π2,π2]{0}

Ans: Correct answer is option D 

Given: cosec1x

Use the properties of inverse trigonometric functions.

The principal value branch of cosec1x is [π2,π2]{0} as cosec1(0)= which is not defined.

Hence, [π2,π2]{0} is the correct answer.


22. If 3tan1x+cot1x=π, then x equals

(A) 0

(B) 1

(C) 1

(D) 12

Ans: Correct answer is option B.

Given: 3tan1x+cot1x=π

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

3tan1x+cot1x=π

2tan1x+tan1x+cot1x=π

2tan1x+π2=π

2tan1x=π2

tan1x=π4

x=1

Hence, x=1.


23. The value of sin1(cos(33π5)) is

(A) 3π5

(B) 7π5

(C) π10

(D) π10

Ans: Correct answer is option D.

Given: sin1(cos(33π5))

Use the properties of inverse trigonometric functions.

sin1(cos(33π5))=sin1(cos(30π+3π5))

sin1(cos(33π5))=sin1(cos(6π+3π5))

sin1(cos(33π5))=sin1(cos(3π5))

sin1(cos(33π5))=sin1(cos(π2+π10))

sin1(cos(33π5))=sin1(sin(π10))

sin1(cos(33π5))=sin1(sin(π10))

sin1(cos(33π5))=π10

Hence, sin1(cos(33π5))=π10.


24. The domain of the function cos1(2x1) is

(A) [0,1]

(B) [1,1]

(C) (1,1)

(D) [0,π]

Ans: Correct answer is option A.

Given: cos1(2x1)

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

Let f(x)=cos1(2x1)

12x11

1+12x1+1

02x2

0x1

Hence, the domain of cos1(2x1) is [0,1].


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

Ans: Correct answer is option A.

Given: f(x)=sin1x1

Use the properties of inverse trigonometric functions.

Let f(x)=sin1x1

As x10 and 1x11,

0x11

1x2

Hence, the domain of f(x)=sin1x1 is [1,2].


26. If cos(sin125+cos1x)=0, then x is equal to

(A) 15

(B) 25

(C) 0

(D) 1

Ans: Correct answer is option B.

Given: cos(sin125+cos1x)=0

Use the properties of inverse trigonometric functions.

cos(sin125+cos1x)=0

sin125+cos1x=cos1(0)

sin125+cos1x=π2

sin125=π2cos1x

sin125=sin1x

x=25

Hence, x=25.


27. The value of sin(2tan1(.75)) is equal to

(A) .75

(B) 1.5

(C) .96

(D) sin1.5

Ans:  Correct answer is option C.

Given: sin(2tan1(.75))

Use the properties of inverse trigonometric functions.

sin(2tan1(.75))=sin(2tan134)

As 2tan1x=sin12x1+x2,

sin(2tan1(.75))=sin(sin12×341+916)

sin(2tan1(.75))=sin(sin1322516)

sin(2tan1(.75))=sin(sin12425)

sin(2tan1(.75))=sin[sin1(0.96)]

sin(2tan1(.75))=0.96

Hence, sin(2tan1(.75))=0.96.


28. The value of cos1(cos3π2) is equal to

(A) π2

(B) 3π2

(C) 5π2

(D) 7π2

Ans:  Correct answer is option A.

Given: cos1(cos3π2)

Use the properties of inverse trigonometric functions.

cos1(cos3π2)=cos1[cos(π+π2)]

cos1(cos3π2)=cos1[cosπ2]

cos1(cos3π2)=cos1[0]

cos1(cos3π2)=π2

Hence, cos1(cos3π2)=π2.


29. The value of the expression 2sec12+sin1(12) is

(A) π6

(B) 5π6

(C) 7π6

(D) 1

Ans:  Correct answer is option A.

Given: 2sec12+sin1(12)

Use the properties of inverse trigonometric functions.

2sec12+sin1(12)=2sec1(secπ3)+sin1(sinπ6)

2sec12+sin1(12)=2.π3+π6

2sec12+sin1(12)=5π6

Hence, 2sec12+sin1(12)=5π6.


30. If tan1x+tan1y=4π5, then cot1x+cot1y equals

(A) π5

(B) 2π5

(C) 3π5

(D) π

Ans:  Correct answer is option A.

Given: tan1x+tan1y=4π5

Use the properties of inverse trigonometric functions.

tan1x+tan1y=4π5

π2cot1x+π2cot1y=4π5

π(cot1x+cot1y)=4π5

cot1x+cot1y=π5

Hence, cot1x+cot1y=π5.


31. If sin1(2a1+a2)+cos1(1a21+a2)=tan1(2x1x2), where a,x[0,1]. then the value of x is

(A) 0

(B) a2

(C) a

(D) 2a1a2

Ans: Correct answer is option D.

Given: sin1(2a1+a2)+cos1(1a21+a2)=tan1(2x1x2)

Use the properties of inverse trigonometric functions.

Given that sin1(2a1+a2)+cos1(1a21+a2)=tan1(2x1x2)

As 2tan1x=sin1(2x1+x2)=cos1(1x21+x2)=tan1(2x1x2),

2tan1a+2tan1a=2tan1x

4tan1a=2tan1x

2tan1a=tan1x

tan12a1a2=tan1x

x=2a1a2

Hence, x=2a1a2.


32. The value of cot[cos1(725)] is

(A) 2524

(B) 257

(C) 2425

(D) 724

Ans: Correct answer is option D.

Given: cot[cos1(725)]

Use the properties of inverse trigonometric functions.

Let cos1(725)=θ

cosθ=725

cotθ=724

cot[cos1(725)]=cotθ

cot[cos1(725)]=724

Hence, cot[cos1(725)]=724.


33. The value of the expression tan(12cos125) is

(A) 2+5

(B) 52

(C) 5+22

(D) 5+2

[Hint:tanθ2=1cosθ1+cosθ]

Ans: Correct answer is option B.

Given: tan(12cos125)

Use the properties of inverse trigonometric functions.

Let θ=12cos125

2θ=cos125

cos2θ=25

As cos2θ=1tan2θ1+tan2θ,

1tan2θ1+tan2θ=25

2+2tan2θ=55tan2θ

(5+2)tan2θ=52

tan2θ=525+2

tan2θ=(52)(52)(5+2)(52)

tan2θ=(52)254

tanθ=±(52)

tanθ=52

tan(12cos125)=52

Hence, tan(12cos125)=52.


34. If |x|1,then 2tan1x+sin1(2x1+x2) is equal to

(A) 4tan1x

(B) 0

(C) π2

(D) π

Ans:  Correct answer is option A.

Given: |x|1,

Use the properties of inverse trigonometric functions.

As 2tan1x=sin1(2x1+x2),

2tan1x+sin1(2x1+x2)=2tan1x+2tan1x

2tan1x+sin1(2x1+x2)=4tan1x

Hence, 2tan1x+sin1(2x1+x2)=4tan1x..


35. If cos1α+cos1β+cos1γ=3π, then α(β+γ)+β(γ+α)+γ(α+β) equals

(A) 0

(B) 1

(C) 6

(D) 12

Ans:  Correct answer is option C.

Given: cos1α+cos1β+cos1γ=3π

Use the properties of inverse trigonometric functions.

Given that cos1α+cos1β+cos1γ=3π

cos1α+cos1β+cos1γ=π+π+π

cos1α=π, cos1β=π, cos1γ=π

α=cosπ, β=cosπ, γ=cosπ

α=1, β=1, γ=1

α(β+γ)+β(γ+α)+γ(α+β)=(1)(11)+(1)(11)+(1)(11)

α(β+γ)+β(γ+α)+γ(α+β)=6

Hence, α(β+γ)+β(γ+α)+γ(α+β)=6.


36. The number of real solutions of the equation 1+cos2x=2cos1(cosx) in [π2,π] is

(A) 0

(B) 1

(C) 2

(D) Infinite

Ans:  Correct answer is option A.

Given: 1+cos2x=2cos1(cosx)

Use the properties of inverse trigonometric functions.

1+cos2x=2cos1(cosx)

2cos2x=2x

2cosx=2x

cosx=x which is not possible for any value of x.

Hence, the number of real solutions of 1+cos2x=2cos1(cosx) in [π2,π] is 0..


37. If cos1x>sin1x, then

(A) 12<x1

(B) 0x<12      

(C) 1x<12

(D) x>0

Ans:  Correct answer is option C.

Given: cos1x>sin1x

: The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

cos1x>sin1x

sin[cos1x]>x

sin[sin11x2]>x

1x2>x

x<1x2

x2<1x2

2x2<1

x2<12

x<±12

As 1x1, 1x<12.

Hence, 1x<12.


38. The principal value of cos1(12) is

Ans:  Given: cos1(12)

Use the properties of inverse trigonometric functions.

Let cos1(12)=x

cosx=12

cosx=cos(π3)

cosx=cos(ππ3)

cosx=cos(2π3)

x=2π3[0,π]

Therefore, the principal value of cos1(12) is 2π3.


39. The value of sin1(sin3π5) is

Ans:  Given: sin1(sin3π5)

Use the properties of inverse trigonometric functions.

sin1(sin3π5)3π5 as 3π5[π2,π2]

So, sin1(sin3π5)=sin1sin(π2π5)

sin1(sin3π5)=sin1sin(2π5)

sin1(sin3π5)=2π5[π2,π2]

Therefore, the value of sin1(sin3π5) is 2π5.


40. If cos(tan1x+cot13)=0, then value of x is

Ans:  Given: cos(tan1x+cot13)=0

Use the properties of inverse trigonometric functions.

cos(tan1x+cot13)=0

tan1x+cot13=cos1(0)

tan1x+cot13=π2

tan1x=π2cot13

tan1x=tan13

x=3

Therefore, x=3.


41. The set of values of sec1(12) is

Ans:  Given: sec1(12)

Use the properties of inverse trigonometric functions.

Let sec1(12)=x

secx=12

The domain of sec1x is R{1,1} and 12R{1,1}

Therefore, sec1(12) has no set of values.


42. The principal value of tan13 is

Ans:  Given: tan13

Use the properties of inverse trigonometric functions.

tan13=tan1(tanπ3)

tan13=π3(π2,π2)

Therefore, the principal value of tan13 is π3.


43. The value of cos1(cos14π3) is

Ans:  Given: cos1(cos14π3)

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

cos1(cos14π3)=cos1[cos(5ππ3)]

cos1(cos14π3)=cos1[cos(π3)]

cos1(cos14π3)=cos1[cos(ππ3)]

cos1(cos14π3)=cos1[cos(2π3)]

cos1(cos14π3)=2π3[0,π]

Therefore, the value of cos1(cos14π3) is 2π3.


44. The value of cos(sin1x+cos1x),|x|1 is

Ans:  Given: |x|1

Use the properties of inverse trigonometric functions.

cos(sin1x+cos1x)=cosπ2

cos(sin1x+cos1x)=0

Therefore, cos(sin1x+cos1x)=0.


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

Ans:  Given: x=32

Use the properties of inverse trigonometric functions.

tan(sin1x+cos1x2)=tan(π4)

tan(sin1x+cos1x2)=1

Therefore, tan(sin1x+cos1x2)=1.


46. If y=2tan1x+sin1(2x1+x2) for all x, then .....<y<.....

Ans:  Given: y=2tan1x+sin1(2x1+x2)

Use the properties of inverse trigonometric functions.

y=2tan1x+sin1(2x1+x2)

As 2tan1x=sin1(2x1+x2),

y=2tan1x+2tan1x

y=4tan1x

As π2<tan1x<π2,

4×π2<4tan1x<4×π2

2π<y<2π

Therefore, 2π<y<2π


47. The result tan1xtan1y=tan1(xy1+xy) is true when value of xy is

Ans:  Given: tan1xtan1y=tan1(xy1+xy)

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

tan1xtan1y=tan1(xy1+xy) is true when xy>1.

Therefore, xy>1.


48. The value of cot1(x) for all xR in terms of cot1x is

Ans:  Given: cot1(x)

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

cot1(x)=πcot1x, xR

Therefore, cot1(x)=πcot1x, xR.


State True or False for the statement in each of the Exercises 49 to 55:

49. All trigonometric functions have inverse over their respective domains.

Ans:  Given: All trigonometric functions have inverse over their respective domains.

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

All the inverse trigonometric functions are restricted over their respective domains.

Therefore, the statement is false.


50. The value of the expression (cos1x)2 is equal to sec2x.

Ans:  Given: The value of the expression (cos1x)2 is equal to sec2x.

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

As cos1x=sec1(1x)secx,

(cos1x)2sec2x

Therefore, the statement is false.


51. The domain of trigonometric functions can be restricted to any one of their branches (not necessarily principal value) in order to obtain their reverse functions.

Ans:  Given: The domain of trigonometric functions can be restricted to any one of their branch (not necessarily principal value) in order to obtain their reverse functions.

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

All the trigonometric functions are restricted over their respective domains to get their inverse functions.

Therefore, the statement is true.


52. The least numerical value, either positive or negative, of angle θ is called the principal value of the inverse trigonometric function.

Ans:  Given: The least numerical value, either positive or negative of angle θ is called the principal value of the inverse trigonometric function.

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

Principal value of any inverse trigonometric function is the least numerical value, either positive or negative of θ

Therefore, the statement is true.


53. The graph of inverse trigonometric functions can be obtained from the graph of their corresponding trigonometric function by interchanging x and y axes.

Ans:  Given: The graph of inverse trigonometric function can be obtained from the graph of their corresponding trigonometric function by interchanging xand yaxes.

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

The domain and the range are interchanged in the graph of inverse trigonometric functions to that of their corresponding trigonometric functions.

Therefore, the statement is true.


54. The minimum value of n for which tan1nπ>π4,nN is valid is 5.

Ans:  Given: The minimum value of n for which tan1nπ>π4, nN is valid is 5.

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, secant, tangent, and cosecant are called inverse trigonometric functions.

tan1nπ>π4

nπ>tanπ4

nπ>1

n>π

n>3.14

So, the value of n is 4.

Therefore, the statement is false.


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

Ans:  Given: The principal value of sin1[cos(sin112)] is π3.

The inverse functions of the basic trigonometric functions sine, cotangent, cosine, 

sin1[cos(sin112)]=sin1[cos(sin1sinπ6)]

sin1[cos(sin112)]=sin1[cosπ6]

sin1[cos(sin112)]=sin1(32)

sin1[cos(sin112)]=sin1(sinπ3)

sin1[cos(sin112)]=π3

Therefore, the statement is true.


In the  NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions), the following sub-topics are covered:

2.1 An Overview

2.2 Fundamental ideas

2.3 Inverse Trigonometric function properties


Students will obtain the complete solution to the questions in the NCERT  Exemplar book after each topic thoroughly with our comprehensive NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions)


NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions) covers a variety of topics that will assist you in preparing for higher learning and Examinations. Class 12 Maths Solutions from NCERT Inverse Trigonometric Functions, their principles, range, domain, and Functions are all covered in Chapter 2.

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FAQs on NCERT Exemplar for Class 12 Maths Chapter-2 (Book Solutions)

1. Why should you refer to NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions)?

The topic that demands some time and effort on the side of the learner is covered in NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions). It is a crucial Chapter in Calculus Mathematics since it clarifies the concept of integrals and their presence. We'll assist you in locating the answers to all of the NCERT exemplar questions in this Chapter. Learners will be able to understand Inverse Trigonometric Functions from NCERT Exemplar Class 12 Maths Chapter 2 answers. Learners can utilise the online webpage of Vedantu to download the PDF of NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions).

2. How will NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions) help you?

The techniques on how to find integrals and what are the ranges and domains of these Inverse Functions are covered in NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions). It is a very interesting Chapter and its concepts build a base that will also help you in solving the later Chapters effectively.

Other themes covered in the NCERT Example Class 12 Maths Chapter 2 solutions include what makes these Functions Inverse, how they act, and many qualities. As a result, NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions) will assist students in solving questions later in the Chapter.

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It's critical to understand sine, cosine, tangent, and their Inverse Trigonometric Functions, as well as their Inverses. This Chapter also will make it much easier to understand the problems by answering the questions straightforwardly and comprehensively. The knowledgeable faculties and instructors of Vedantu have answered the questions in the most straightforward manner possible. Students will be able to understand the topic and problems better with the help of answers.

NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions) are comprehensive, with additional steps and formulas at each stage. This will make the answer more specific for every student.

4. What will students learn inNCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions)?

Understanding and mastering this Chapter will aid one in achieving a higher grade in school, boards, and entrance tests.

Students will obtain complete answers to the questions in the NCERT book after each topic with our comprehensive Class 12 Maths NCERT Exemplar Solutions Chapter 2. With the use of questions and solved Examples, the topics presented in NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions)  will assist in a lot more complete comprehension of Inverse Trigonometric Functions.

It is critical to grasp the properties of each Inverse Trigonometric function to solve problems and comprehend theorems and principles. As a result, the relationship and properties of the Inverse Trigonometric Functions cot, tan, sine, sec, cosec, cos, and sec are covered.

5. What are the important topics that will be covered in NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions)?

NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions) contains important subjects to cover.

In the NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions), the importance of understanding Inverse Trigonometric Functions and their properties will be discussed. It explains the fundamentals of Inverse Trigonometric Functions. Students will learn about the characteristics and graphical representations of Inverse Trigonometric Functions in NCERT Exemplar for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions (Book Solutions).

In Chapter 2 of the NCERT Example solutions for Class 12 Maths, students will study the fundamental ideas of Trigonometric Functions such as sine, cosine, and cosec, as well as their characteristics.