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RD Sharma Solutions Class 12 Mathematics Chapter 3 VSA

RD Sharma Solutions Class 12 Mathematics Chapter 3 VSA

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 12:37 PM IST

RD Sharma has been the choice of many students and teachers all over the country. Their simple easy to understand solutions make it very beneficial for students to learn the concepts quickly. Many teachers use these materials for lectures and to set up question papers. As these contain numerous examples, students can be thorough with their preparation and leave no problem unsolved.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

Inverse Trigonometric Functions Excercise: VSAQ

RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Background wave

Answer:
C=a+b2
Hint:
Find differentiation of f(x) and apply Rolle’s Theorem.
Given:
f(x)=Ax2+Bx+C and f(a)=f(b)
Solution:
f(a)=f(b)
⇒Aa2+Ba+C=Ab2+Bb+C⇒Aa2+Ba=Ab2+Bb⇒A(a2−b2)+B(a−b)=0⇒A(a−b)(a+b)+B(a−b)=0⇒(a−b)[A(a+b)+B]=0⇒A(a+b)+B=0 and a−b=0

⇒A=−Ba+b and a−b
⇒A=−Ba+b [∵a≠b]
Now, f(x)=Ax2+Bx+C
f′(x)=2xA+B
f′(C)=2AC+B
By Rolle’s Theorem,
⇒ f′(C)=0
⇒ 2AC+B=0
⇒ C=−B2A
⇒ C=−B2(−Ba+b) [∵A=−Ba+b]
⇒ C=−B2×a+b(−B)
⇒ C=a+b2

Inverse Trigonometric Function Exercise Very Short Answer Question 2

Answer:
Statement of Rolle’s Theorem
Hint:
You must know about Rolle’s Theorem.
Given:
Rolle’s Theorem
Solution:
Rolle’s Theorem states that if f(x) is continuous in the closed interval [a,b] and differentiable on open interval (a,b) such that f(a)=f(b) then f′(x)=0 for some a<x<b.


Inverse Trigonometric Function Exercise Very Short Answer Question

Answer:
Statement of mean value theorem.
Hint:
You must know about mean value theorem.
Given:
Lagrange’s mean value theorem.
Solution:
Let be a function that satisfies the following hypothesis.
  1. f(x) is continuous on the closed interval [a,b].
  2. f(x) is differentiable on the open interval (a,b).
Then there is a value in (a,b) such that
f′(c)=f(b)−f(a)b−a
Or equivalently
f(b)−f(a)=f′(c)(b−a)

Inverse Trigonometric Function Exercise Very Short Answer Question 4

Answer:
n=3
Hint:
Find differentiate of f(x) and then apply Rolle’s Theorem.
Given:
f(x)=2x(x−3)n,xϵ[0,23]and value of c is 34.
Solution:
According to Rolle’s Theorem
f(a)=f(b)⇒f(0)=f(23)⇒0=2⋅23(23−3)n
⇒(23−3)n=0
Now, f(x)=2x(x−3)n
Differentiate with respect to x,
f′(x)=2(x−3)n+2x⋅n(x−3)n−1 [∵d(uv)dx=udvdx+vdudx]
Applying Rolle’s Theorem,
f′(c)=2(c−3)n+2c⋅n(c−3)n−1
f′(c)=0
2(c−3)n+2c⋅n(c−3)n−1=0
⇒2(34−3)n+2⋅34⋅n(34−3)n−1=0 [∵c=34]
⇒(34−3)n[2(34−3)+3n2]=0
⇒(34−3)n−1=0 or 2(3−124)+3n2=0
⇒ 3−12+3n2=0
3−12+3n=0
3n=9
n=3

Inverse Trigonometric Function Exercise Very Short Answer Question 5

Answer:
c=5
Hint:
Find differentiability of f(x) and then apply the formula of f′(c).
Given:
f(x)=x2−4,xϵ[2,3]
Solution:
f(x)=x2−4
f(x)=(x2−4)12
Differentiate,
⇒f′(x)=12(x2−4)12−1⋅2x [∵ddx(xn)=n(x)n−1]
⇒f′(x)=12(x2−4)−12⋅2x⇒f′(x)=12x2−4⋅2x⇒f′(x)=xx2−4⇒f′(c)=cc2−4
Use mean value theorem,
f′(c)=f(b)−f(a)b−a⇒cc2−4=5−03−2⇒cc2−4=5⇒c=5(c2−4)⇒c2=5(c2−4)
⇒c2=5c2−20
⇒4c2=20
⇒c2=5
⇒c=±5 [∵xϵ[2,3]]
⇒c=5

Inverse Trigonometric Functions exercise Very short answer question 1

Answer: π3
Given: sin−1⁡(−32)+cos−1⁡(−12)
Hint: sin−1⁡(−x)=−sin−1⁡x,x∈(−1,1)
cos−1⁡(−x)=π−cos−1⁡x,x∈(−1,1)
Solution:
sin−1⁡(−32)+cos−1⁡(−12)=−sin−1⁡(32)+π−cos−1⁡(12)=sin−1⁡(sin⁡π3)+π−cos−1⁡(cos⁡π3)=−π3+π−π3=π3

Inverse Trigonometric Functions exercise Very short answer question 2

Answer: π
Given: sin−1⁡x,x∈(−1,1)
Hint: The maximum value of sin−1⁡x in x∈(−1,1) is at 1 .
Solution:
sin−1⁡(1) maximum value is,
=sin−1⁡(sin⁡π2)
=π2
Again the minimum value is at -1
So,
sin−1⁡(−1)=−sin−1⁡(1)=−sin−1⁡(π2)=−π2
The difference between the minimum value, maximum value is
(π2−(−π2))=π

Inverse Trigonometric Functions exercise Very short answer question

Answer:
Given: sin−1⁡x+sin−1⁡y+sin−1⁡z=3π2,x+y+z=?
Hint: The maximum value in the range of sin−1 is π2
Solution:
sin−1⁡x+sin−1⁡y+sin−1⁡z=π2+π2+π2
Here sum of three inverse of sin⁡x is terms π2, i.e. every sin inverse function is equal to π2.
sin−1⁡x=π2,sin−1⁡y=π2,sin−1⁡z=π2x=sin⁡π2,y=sin⁡π2,z=sin⁡π2x=1,y=1,z=1x+y+z=1+1+1=3

Inverse Trigonometric Functions exercise Very short answer question 4

Answer:π−2tan−1⁡x
Given:
x>1,sin−1⁡(2x1+x2)
Hint:
2tan−1⁡x=π−sin−1⁡(2x1+x2)forx>1
Solution:
sin−1⁡(2x1+x2)=π−2tan−1⁡x

Inverse Trigonometric Functions exercise Very short answer question 5

Answer: −2tan−1⁡x
Given:
x<0,cos−1⁡(1−x21+x2)
Hint:
1−tan2⁡x1+tan2⁡x=cos⁡2x
Solution:
Let, x=tan⁡y
cos−1⁡(1−x21+x2)=cos−1⁡(1−tan2⁡y1+tan2⁡y)=cos−1⁡(cos⁡2y)=2y−(1)
The value of x is negative, so let x=-a where a>0 .
−a=tan⁡yy=tan−1⁡(−a)

Now,
cos−1⁡(1−x21+x2)=2y Using (1)
=2tan−1⁡(−a)=2tan1⁡x[∴x=−a]

Inverse Trigonometric Functions exercise Very short answer question 6

Answer: π2
Given:
tan−1⁡x+tan−1⁡(1x)  \  for  x>0
Hint:
tan−1⁡x+tan−1⁡(1x)=tan−1⁡(x+1x1−x1x),x>0
Solution:
tan−1⁡x+tan−1⁡(1x)=tan−1⁡(x2+10)=tan−1⁡(∞)=tan−1⁡(tan⁡π2)=π2

Inverse Trigonometric Functions exercise Very short answer question 7

Answer:−π2
Given:
tan−1⁡x+tan−1⁡(1x) for x<0
Hint:
tan−1⁡x+tan−1⁡y=tan−1⁡(x+y1−xy) when x<0,1x<0
Solution:
$Let \ \ x=-y, y>0$
Then,
tan−1⁡x+tan−1⁡(1x)=tan−1⁡(−y)+tan−1⁡(−1y)=−(tan−1⁡y+tan−1⁡1y)=−tan−1⁡(y+1y1−yy1),y>0=−tan−1⁡(y2+10)=−tan−1⁡(∞)=−tan−1⁡(tan⁡π2)=−π2

Inverse Trigonometric Functions exercise Very short answer question 8

Answer: π
Given:
cos−1⁡(cos⁡2π3)+sin−1⁡(sin⁡2π3)
Hint:
Range of sin⁡(−π2,π2);π3∈(−π2,π2)
Range of cos⁡[0,π],2π3∈(0,π)
Solution:
cos−1⁡(cos⁡2x3)+sin−1⁡(sin⁡2x3)=cos−1⁡(cos⁡2x3)+sin−1⁡[sin⁡(2x3)]=2π3+π−2π3=π

Inverse Trigonometric Functions exercise Very short answer question 9

Answer: 0
Given:
−1<x<0,sin−1⁡(2x1+x2)+cos−1⁡(1−x21+x2)
Hint: Try to convert the sin and cos function into tan .
Solution:
 Let, x=−tan⁡y,0<y<π2
sin−1⁡(2x1+x2)+cos−1⁡(1−x21+x2)=sin−1⁡(−2tan⁡y1+tan2⁡y)+cos−1⁡(1−tan2⁡y1+tan2⁡y)=sin−1⁡{−sin⁡(2y)}+cos−1⁡{cos⁡(2y)}=−sin−1⁡{sin⁡2y}+cos−1⁡{cos⁡(2y)}=−2y+2y=0

Inverse Trigonometric Functions exercise Very short answer question 10

Answer: 1x2+1
Given: sin⁡(cot−1⁡x)
Hint: sin⁡(cot−1⁡x)=sin⁡(tan−1⁡1x)
tan−1⁡x=sin⁡(x1+x2)
Solution:
cot−1⁡x=tan−1⁡1x
sin⁡(cot−1⁡x)=sin⁡(tan−1⁡1x)
=sin⁡[sin−1⁡(1xx2+1x)]
=sin⁡(sin−1⁡1x2+1)
=1x2+1∴(sin⁡(sin−1⁡x)=x)

Inverse Trigonometric Functions exercise Very short answer question 11

Answer: 2π3
Given:
cos−1⁡(12)+2sin−1⁡(12)
Hint: The range of sin⁡is\((−π2,π2),  range  of  cos⁡is  \((0,π)
Solution:

cos−1⁡(12)+2sin−1⁡(12)=cos−1⁡(cos⁡π3)+2sin−1⁡(sin⁡π6)=π3+2(π6)=π3+π3=2π3

Inverse Trigonometric Functions exercise Very short answer question 12

Answer:(−π2,π2)
Given:
Range of tan−1⁡x
Hint: Range of tan function should be all real number.
Solution:
tan−1⁡x=(−π2,π2)

Inverse Trigonometric Functions exercise Very short answer question 1

Answer: 100∘
Given:
cos−1⁡(cos⁡1540∘)
Hint: Try to separate the degree into two parts.
Solution:
cos−1⁡(cos⁡1540∘)=cos−1⁡{cos⁡(1440∘+100∘)}=cos−1⁡{cos⁡(100∘)}∴[cos⁡(4π+100∘)=cos⁡100∘]=100∘

Inverse Trigonometric Functions exercise Very short answer question 14

Answer: 60∘
Given:
sin−1⁡(sin⁡(−600∘))
Hint:
sin⁡x=sin⁡(π−x)
Solution:
sin−1⁡{sin⁡x}=x

sin−1⁡{sin⁡(600∘)}=sin−1⁡{sin⁡(720∘−600∘)}
=sin−1⁡{sin⁡(120∘)}=sin−1⁡{sin⁡(180∘−120∘)}=sin−1⁡{sin⁡(60∘)}=60∘ 

Inverse Trigonometric Functions exercise Very short answer question 15

Answer: 79
Given:
cos⁡(2sin−1⁡13)
Hint: Try to solve the bracket portion
i.e. sin−1 function.
Solution:
Let  ,y=sin−1⁡13Then  ,sin⁡y=13
Now,
cos⁡y=1−sin2⁡y,cos⁡y=1−19=89=223cos⁡(2sin−1⁡13)=cos⁡2y
=cos2⁡y−sin2⁡y[∴cos⁡2x=cos2⁡x−sin2⁡x]=(223)2−(13)2=89−19=79

Inverse Trigonometric Functions exercise Very short answer question 16

Answer: 70∘
Given:
sin−1⁡(sin⁡1550∘)
Hint:
sin−1⁡(sin⁡x)=x
Solution:
We, know that
sin−1⁡(sin⁡x)=x
Now,
sin−1⁡(sin⁡1550∘)=sin−1⁡{sin⁡(1620∘−1550∘)}=sin−1⁡(sin⁡70∘)=70∘

Inverse Trigonometric Functions exercise Very short answer question 17

Answer: 110
Given:
sin⁡(12cos−1⁡45)
Hint:
sin−1⁡(sin⁡x)=x
Solution:
cos−1⁡x=2tan−1⁡1−x1+xtan−1⁡x=sin−1⁡x1+x2∴sin⁡(12cos−1⁡45)=sin⁡(12(2tan−1⁡1−451+45))=sin⁡(tan−1⁡1595)=sin⁡(tan−1⁡13)
=sin⁡{sin−1⁡(121+19)}=sin⁡{sin−1⁡110}=110

Inverse Trigonometric Functions exercise Very short answer question 18

Answer: 35
Given:
sin⁡(tan−1⁡34)
Hint: Try to solve tan function first.
Solution:
tan−1⁡x=sin−1⁡x1+x2
sin⁡(tan−1⁡34)=sin⁡{sin−1⁡(341+916)}
=sin⁡{sin−1⁡(2454)}
=sin⁡{sin−1⁡35}=35

Inverse Trigonometric Functions exercise Very short answer question 19

Answer: π
Given:
cos−1⁡(tan⁡3π4)
Hint: Try to solve tan function first.
∴tan⁡(π−x)=tan⁡x
Solution:
cos−1⁡(tan⁡3π4)=cos−1⁡{−tan⁡(π−3π4)}=cos−1⁡{tan⁡(−π4)}=cos−1⁡{−tan⁡π4}=cos−1⁡(−1)=cos−1⁡(cos⁡π)=π

Inverse Trigonometric Functions exercise Very short answer question 20

Answer:
12
Given
cos⁡(2sin−1⁡12)
Hint: Try to solve sin function.
Solution:
cos⁡(2sin−1⁡12)=cos⁡(2×π6)=cos⁡(π3)=12

Inverse Trigonometric Functions exercise Very short answer question 21

Answer: 20∘
Given: cos−1⁡(cos⁡350∘)−sin−1⁡(sin⁡350∘)
Hint: [sin⁡(360∘−x)=−sin⁡x,cos⁡(360∘−x)=cos⁡x]
Solution:
cos−1⁡(cos⁡350∘)−sin1⁡(sin⁡350∘)
=cos−1⁡{cos⁡(360∘−350∘)}−sin−1⁡{sin⁡(360∘−350∘)}
=cos−1⁡{cos⁡(10∘)}−sin−1⁡{sin⁡(−10∘)}
=10∘−(−10∘)
=20∘

Inverse Trigonometric Functions exercise Very short answer question 22

Answer: 45
Given: cos2⁡(12cos−1⁡35)
Hint: cos⁡2x=2cos2⁡x−1
Solution:
 Let, y=cos−1⁡(35)
cos⁡y=35
 Now, cos2⁡(12cos−1⁡35)=cos2⁡(12y)
=cos⁡y+12[∴cos⁡2x=2cos2⁡x−1]
=z5+12
=852
=45

Inverse Trigonometric Functions exercise Very short answer question 2

Answer: 1
Given:
tan−1⁡x+tan−1⁡y=π4
Hint: Try to separate the tan function into RHS, So that the variables gets free.
Solution:
tan−1⁡x+tan−1⁡y=tan−1⁡(x+y1−xy)
Now,
tan−1⁡x+tan−1⁡y=π4tan−1⁡(x+y1−xy)=π4x+y1−xy=tan⁡π4x+y1−xy=1x+y=1−xyx+y+xy=1

Inverse Trigonometric Functions exercise Very short answer question 24

Answer:2π−6
Given: cos−1⁡(cos⁡6)
Solution:
cos−1⁡(cos⁡6)=cos−1⁡{cos⁡(2π−6)}=2π−6

Inverse Trigonometric Functions exercise Very short answer question 25

Answer: 7π18
Given: sin−1⁡(cos⁡π9)
Hint: cos⁡x=sin⁡(π2−x)
Solution:
sin−1⁡(cos⁡π9)=sin−1⁡{sin⁡(π2−π9)}=sin−1⁡{sin⁡(7π18)}=7π18

Inverse Trigonometric Functions exercise Very short answer question 26

Answer: 1
Given: sin⁡{π3−sin−1⁡(−12)}
Hint: Try to solve the sin−1 function.
Solution:
sin⁡{π3−sin−1⁡(−12)}=sin⁡{π3−(−π6)}=sin⁡{π3+π6}=sin⁡π2=1

Inverse Trigonometric Functions exercise Very short answer question 27

Answer:−π4
Given: tan−1⁡{tan⁡(15π4)}
Hint: Try to solve the tan function first.
∴tan⁡(4π−x)=−tan⁡x
Solution:
tan−1⁡{tan⁡(15π4)}=tan−1⁡{tan⁡(4π−π4)}=tan−1⁡{tan⁡(π4)}=tan1⁡{tan⁡(−π4)}=−π4

Inverse Trigonometric Functions exercise Very short answer question 28

Answer: π
Given: 2sin−1⁡12+cos−1⁡(−12)
Hint: Try to solve the sin−1⁡12  &  cos−1⁡12 and add them.
Solution:
2sin−1⁡12+cos−1⁡(−12)=sin−1⁡2×121−(12)2+cos−1⁡(−12)=sin−1⁡32+cos−1⁡(−12)
=sin−1⁡(sin⁡π3)+cos−1⁡(cos⁡2π3)=π3+2π3=π

Inverse Trigonometric Functions exercise Very short answer question 29

Answer:π4
Given:
Hint: tan−1⁡x−tan−1⁡y=tan−1⁡(x−y1+xy)
Solution:
tan−1⁡ab−tan−1⁡(a−ba+b)=tan−1⁡(ab−(a−b)a+b1+ab(a−ba+b))=tan−1⁡(a2+ab−ab+b2b(a+b)ab+b2−ab+a2b(a+b))=tan−1⁡(1)=tan−1⁡(tan⁡π4)∴(tan⁡π4=1)=π4

Inverse Trigonometric Functions exercise Very short answer question 0

Answer: 3π4
Given: cos−1⁡(cos⁡5π4)
Hint: Try to solve cos function first.
Solution:
cos−1⁡(cos⁡5π4)≠5π4  as  5π4 doesn’t lie in between 0 and 5π
We, have
cos−1⁡(cos⁡5π4)=cos−1⁡{cos⁡(2π−3π4)}
=cos−1⁡{cos⁡(3π4)}=3π4

Inverse Trigonometric Functions exercise Very short answer question 1

Answer: 2sin−1⁡x=sin−1⁡(2x1−x2)
Given: To prove that
sin−1⁡(2x1−x2)=2sin−1⁡x
Hint:x=sin⁡a
Solution:
LHS=sin−1⁡2x1−x2 Putting
x=sin⁡a,weget=sin−1⁡(2sin⁡a1−sin2⁡a)=sin−1⁡(2sin⁡acos⁡a)=sin−1⁡(sin⁡2a)=2a=2sin−1⁡x[x=sin⁡a]
∴sin−1⁡(2x1−x2)=2sin−1⁡x

Inverse Trigonometric Functions exercise Very short answer question 2

Answer: 2π5
Given:
sin−1⁡(sin⁡3π5)
Hint:
sin−1⁡(sin⁡x)=x
Solution:
sin−1⁡(sin⁡3π5)=sin−1⁡{sin⁡(π−3π5)}:[(π−3π5)∈[−π2,π2]]
=sin−1⁡(sin⁡2π5)=2π5

Inverse Trigonometric Functions exercise Very short answer question

Answer: 3
Given:
tan−1⁡3+cot−1⁡x=π2
Hint:
tan−1⁡x+cot−1⁡x=π2
Solution:
tan−1⁡(3)+cot−1⁡x=π2tan−1⁡(3)=π2−cot−1⁡xtan−1⁡(3)=tan−1⁡x
x=3

Inverse Trigonometric Functions exercise Very short answer question 4

Answer:
x=13
Given: sin−1⁡(13)+cos−1⁡x=π2
Hint:
sin−1⁡x+cos−1⁡x=π2
Solution:
\sin−1⁡(13)+cos−1⁡x=π2sin−1⁡(13)=π2−cos−1⁡xsin−1⁡(13)=sin−1⁡x
x=12

Inverse Trigonometric Functions exercise Very short answer question 5

Answer:−π2
Given: sin−1⁡(13)−cos−1⁡(−13)
Hint: cos−1⁡x+sin−1⁡x=π2 & cos−1⁡(−x)=π−cos−1⁡x
Solution:
sin−1⁡(13)−cos−1⁡(−13)=sin−1⁡(13)−[π−cos−1⁡(13)]=sin−1⁡(13)+cos−1⁡(13)−π=π2−π=−π2

Inverse Trigonometric Functions exercise Very short answer question 6
Answer:12

Given:
4sin−1⁡x+cos−1⁡x=π
Hint:
sin−1⁡x+cos−1⁡x=π2
Solution:
∴4sin−1⁡x+cos−1⁡x=π4sin−1⁡x+π2−sin−1⁡x=π[∴sin−1⁡x+cos−1⁡x=π2]3sin−1⁡x=π2sin−1⁡x=π6x=sin⁡π6x=12

Inverse Trigonometric Functions exercise Very short answer question 7

Answer: −π2
Given:
x<0,y<0,xy=1
tan−1⁡x+tan−1⁡y=?
Hint:
tan−1⁡x+tan−1⁡y=tan−1⁡(x+y1−xy)
Solution:
x<0,y<0, such that xy=1 Let, x=−a,y=−b
∴tan−1⁡x+tan−1⁡y=tan−1⁡(x+y1−xy)
=tan−1⁡(−a−a1−1)=tan−1⁡(−∞)=tan−1⁡{tan⁡(−π2)}=−π2

Inverse Trigonometric Functions exercise Very short answer question 8

Answer:−π3
Given:
sin−1⁡(−32)
Hint: (−π2,π2) is the range of the principal value branch of inverse sine function.
Solution:
Let  y=sin−1⁡(−32)
Then,
sin⁡y=−32=sin⁡(−π3)
y=−π3∈[−π2,π2]
∴sin−1⁡(−32)=−π3


Inverse Trigonometric Functions exercise Very short answer question 9

Answer: −π6
Given:
sin−1⁡(−12)
Hint:
(−π2,π2) is the range of the principal value branch of inverse sine function.
Solution:
 Let y=sin−1⁡(−12)sin⁡y=−12=sin⁡(−π6)y=−π6∈[−π2,π2]
Here,
[−π2,π2] is range of principal value of sin
sin−1⁡(−12)=−π6

Inverse Trigonometric Functions exercise Very short answer question 40

Answer: π
Given:
cos−1⁡(cos⁡2π3)+sin−1⁡(sin⁡2π3)
Hint:
(−π2,π2) is the range of the principal value branch of sine range of cos⁡(0,π)
Solution:
We have,
cos−1⁡(cos⁡2π3)+sin−1⁡(sin⁡2π3)=cos−1⁡(cos⁡2π3)+sin−1⁡(sin⁡π3)=cos−1⁡(cos⁡2π3)+sin−1⁡(sin⁡π3)=2π3+π3=π

Inverse Trigonometric Functions exercise Very short answer question 41

Answer: 512
Given:
tan⁡(2tan−1⁡15)
Hint:
Use formula
Solution:
tan⁡(2tan−1⁡15)=tan⁡(tan−1⁡2×(15)1−(15)2)=tan⁡(tan−1⁡252425)=512

Inverse Trigonometric Functions exercise Very short answer question 42

Answer: 11π12
Given:
tan−1⁡1+cos−1⁡(−12)
Hint:
Principal range of cos⁡(0,π)
Solution:
tan−1⁡(1)+cos−1⁡(−12)=tan−1⁡(tan⁡π4)+cos−1⁡(cos⁡2π3)=π4+2π3=11π3

Inverse Trigonometric Functions exercise Very short answer question 43

Answer: π3
Given:tan−1⁡[2sin⁡(2cos−1⁡32)]
Hint: Try to solve cos−1 and then sin function.
Solution:
tan−1⁡[2sin⁡(2cos−1⁡32)]=tan−1⁡{2sin⁡[cos−1⁡2(32)2−1]=tan−1⁡[2sin⁡(cos−1⁡12)]=π3

Inverse Trigonometric Functions exercise Very short answer question 44

Answer: π2
Given
:tan−1⁡3+cot−1⁡3
Hint:
tan−1⁡x+cot−1⁡x=π2
Solution:
Using the hint we are able to solve this,
tan−1⁡3+cot−1⁡3=π2

Inverse Trigonometric Functions exercise Very short answer question 45

Answer: 40∘
Given: cos−1⁡(cos⁡680∘)
Hint: Try to solve cos⁡680∘ first, then separate and solve.
Solution:
cos−1⁡(cos⁡680∘)=cos−1⁡[cos⁡(720∘−680∘)]=cos−1⁡(cos⁡40∘)=40∘

Inverse Trigonometric Functions exercise Very short answer question 46

Answer: 2π5
Given: sin−1⁡(sin⁡3π5)
Hint: Try to solve (sin⁡3π5) first.
Solution:
sin−1⁡(sin⁡3π5)=sin−1⁡[sin⁡(π−2π5)]=sin−1⁡(sin⁡2π5)=2π5

Inverse Trigonometric Functions exercise Very short answer question 47

Answer: ϕ
Given:
sec−1⁡(12)
Hint: The range of sec is (-1, 1) .
Solution:
The value of sec−1⁡(12) is undefined as it is outside the range,
R-(-1, 1)

Inverse Trigonometric Functions exercise Very short answer question 48

Answer: 2π3
Given: cos−1⁡[cos⁡(4π+2π3)]
Hint: Try to solve cos⁡(14π3) function.
Solution:
cos−1⁡(cos⁡14π3)=cos−1⁡[cos⁡(4π+2π3)]
=cos−1⁡(cos⁡2π3)=2π3

Inverse Trigonometric Functions exercise Very short answer question 49

Answer: 0
Given: cos⁡(sin−1⁡x+cos−1⁡x),|x|≤1
Hint: |x|≤1,x∈[−1,1]
Solution:
|x|≤1±x≤1x≤1  or  x≥−1x∈[−1,1]
Now,
cos⁡(sin−1⁡x+cos−1⁡x)=cos⁡(π2)=0 

Inverse Trigonometric Functions exercise Very short answer question 50

Answer: 1
Given: tan⁡(sin−1⁡x+cos−1⁡x2) when x=32
Hint: substitute x value into the function.
Solution:
tan⁡(sin−1⁡x+cos−1⁡x2)=tan⁡(sin−1⁡(32)+cos−1⁡(32)2)∴sin−1⁡x+cos−1⁡x=π2
Hence,
tan⁡(sin−1⁡x+cos−1⁡x2)=tan⁡[(π/4)=1

Inverse Trigonometric Functions exercise Very short answer question 51

Answer: π3
Given: sin−1⁡[cos⁡(sin−1⁡12)]
Hint: Try to solve sin−1 function and its cos function.
Solution:
sin−1⁡[cos⁡(sin−1⁡12)]=sin−1⁡{cos⁡[sin−1⁡(sin⁡π3)]}=sin−1⁡[cos⁡(π3)]=sin−1⁡[cos⁡(π3)]=sin−1⁡(12)=sin−1⁡(sin⁡π3)=π3

Inverse Trigonometric Functions exercise Very short answer question 52

Answer: ∅
Given:cosec−1(32)
Hint: Range of principal value of cosec is (−1,1)
Solution:
Range of principal value of is undefined as it’s outside the range
(i.e.,\(R−(−1,1)⟩

Inverse Trigonometric Functions exercise Very short answer question 53

Answer: −π−cot−1⁡x
Given
tan−1⁡(1x) for x<0 in terms of $\cot ^{-1} x$
Hint:
tan−1⁡(1x)=tan−1⁡(−1x) for x<0
Solution:
tan−1⁡(1x)=−tan−1⁡(1x)=−cot−1⁡x=−(π−cot−1⁡x)=−π+cot−1⁡x

Inverse Trigonometric Functions exercise Very short answer question 54

Answer: π−cot−1⁡x
Given:
cot−1⁡(x)  for  all  x∈R in terms of cot−1⁡x
Hint:
cot−1⁡(−x)=π−cot−1⁡x
Solution:
The value of cot−1⁡(−x)=π−cot−1⁡x\) for all x∈Rin terms of cot−1⁡(x) is,
=π−cot−1⁡(x)

Inverse Trigonometric Functions exercise Very short answer question 55

Answer: 32
Given:
cos⁡(tan−1⁡x+cot−1⁡x3)when  x=−13
Hint:
tan−1⁡x+cot−1⁡x=π2
Solution:
cos⁡(tan−1⁡x+cot−1⁡x3)=cos⁡(π6)=32

Inverse Trigonometric Functions exercise Very short answer question 56

Answer: 3
Given:
cos⁡(tan−1⁡x+cot−1⁡3)=0
Hint:
tan−1⁡y+cot−1⁡y=π2
Solution:
cos⁡(tan−1⁡x+cot−1⁡3)=cos⁡(π2)tan−1⁡x+cot−1⁡3=π2x=3

Inverse Trigonometric Functions exercise Very short answer question 57

Answer: 5π6
Given:
2sec−1⁡2+sin−1⁡(12)
Hint: Try to convert or find the values of
sin−1⁡(12),sec−1⁡(2)
Solution:
2sec−1⁡2+sin−1⁡(12)=2sec1⁡(sec⁡π3)+sin−1⁡(sin⁡π6)=2×π3+π6=5π6

Inverse Trigonometric Functions exercise Very short answer question 58

Answer: 25
Given:
cos⁡(sin−1⁡25+cos−1⁡x)=0
Hint: Try to convert the separate sin , cos function. So, that the variable get free.
Solution:
cos⁡(sin−1⁡25+cos−1⁡x)=0cos⁡(sin−1⁡25+cos−1⁡x)=cos⁡(π2)sin−1⁡25+cos−1⁡x=π2x=25cgfy

Inverse Trigonometric Functions exercise Very short answer question 59

Answer: π6
Given:
cos−1⁡(cos−1⁡3π6)
Hint: To solve cos function we can get the solution.
Solution:
cos−1⁡(cos⁡13π6)=cos−1⁡[cos⁡(2π+π6)]=cos−1⁡(cos⁡(π6))=π6

Inverse Trigonometric Functions exercise Very short answer question 60

Answer: π8
Given
tan−1⁡(tan⁡9π8)
Hint: To solve tan function we can get the solution.
Solution:
tan−1⁡(tan⁡9π8)=tan−1⁡(tan⁡(π+π8))=tan−1⁡(tan⁡(π8))=π8

Inverse Trigonometric Functions exercise Very short answer question 61

Answer: −π2
Given: tan−1⁡3−cot−1⁡(−3)
Hint: Range of
tan⁡−1  is  (−π2,π2);π3∈(−π2,π2)
Range of
cot−1⁡is(0,π);5π6∈[0,π]
Solution:
tan−1⁡(3)−cot−1⁡(−3)=tan−1⁡[tan⁡(π3)]−cot−1⁡(cot⁡5π6)=π3−5π6=−π2

Inverse Trigonometric Functions exercise Very short answer question 62

Answer: −π8
Given:
sin−1⁡[(sin−17π8)]
Hint:
sin⁡(sin−1⁡θ)=θ  if  −π2≤θ≤π2
Solution:
sin−1⁡(sin−17π8)sin−1⁡(−sin⁡17π8)=sin−1⁡[−sin⁡(2π+π8)]=sin−1⁡(−sin⁡π8)=sin−1⁡(sin−π8)=−π8

Inverse Trigonometric Functions exercise Very short answer question 63

Answer: 3π4
Given: cot−1⁡2  and  cot−1⁡3 , then third angle.
Hint: The sum of tree angles of a triangle is π
Solution:
tan−1⁡13+tan−1⁡12+tan−1⁡x=πtan−1⁡(13+12)1−16+tan−1⁡x=π

RD Sharma Class 12th VSAQ is based on Inverse Trigonometric Functions. To understand this chapter, students need to have a good idea about trigonometry. However, even if you don't have enough experience with trigonometry, you can still learn this chapter as the solution given by Career360 is for students to understand the chapter from scratch. RD Sharma Class 12th VSAQ can be considered small basic questions that the students can quickly go through.

In RD Sharma Class 12th Chapter 3 VSAQ, you will first evaluate the inverse functions and solve basic sums. This can be solved solely through fundamental operations. The higher-level questions which carry more marks are continued in the next section of this chapter.

There are a total of 62 VSAQs in this chapter which might seem quite overwhelming to students. Still, it is nothing to worry about as Class 12 RD Sharma Chapter 3 VSAQ Solution by Career360 has provided the best solutions which you can refer to and efficiently complete this section.VSAQs, although they carry fewer marks in exams but are essential where the weightage is concerned. It is through these VSAQs that you can learn crucial concepts for lengthy and complex problems. This is why you should always practice every one of them.

Practice, Planning, and Execution are the three factors that define an ideal Math student. As trigonometry consists of hundreds of sums, it is best to plan and divide those questions into parts. This method will be easier as you have to solve several sums per day while learning their concepts. Dividing up your work is very beneficial as it lessens the burden and gives the best output.

As Maths is such a vast subject getting a hold of all the concepts is pretty hard. Teachers can’t explain all the sums from the book, so you need to study these concepts beforehand to reduce the work needed afterward.

As far as trigonometry and inverse trigonometry are concerned, they contain hundreds of sums that are quite a lot for students to practice. This is where RD Sharma Class 12th Chapter 3 VSAQ shines. It includes all essential concepts with solutions for Inverse Trigonometry that can be accessible by everyone for free.

A majority of students have joined the team and started learning from Career360's material. You can also avail of this opportunity and download the material for free. This has excellent solutions which will help you score good marks in exams. So stay on top of your class and excel in trigonometric concepts with the help of RD Sharma Class 12 Chapter 3 VSAQ.

Chapter-wise RD Sharma Class 12 Solutions

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Frequently Asked Questions (FAQs)

1. Does this material have all chapters?

Yes, Career360 offers all the chapters on their website, available for free and accessible by students. You can learn from these solutions to score good marks in exams. For example, to get more insight on Inverse Trigonometry, you can download Class 12 RD Sharma Chapter 3 VSAQ Solution.

2. What are the applications of trigonometry?

These are some of the applications of trigonometry:

  • Oceanography - To calculate tide heights.

  • Calculus

  • Physics - Sine and Cos functions are used as wave representations. 

  • Height calculation - To Measure the height of buildings or mountains.

3. Who invented Inverse Trigonometry?

Daniel Bernoulli introduced inverse trigonometry. This concept was used to determine the angle if two sides of a triangle are given. For more information on Inverse Trigonometry, check out RD Sharma Class 12 Solutions Inverse Trigonometric Functions VSAQ

4. Are NCERT and RD Sharma books similar?

NCERT books for maths are good for general preparation, but RD Sharma books are relatively better to dive deep into the concepts. They are widely considered by teachers and students all over the country and are excellent means of studying for exams. To find the RD Sharma material for Inverse Trigonometry, you can study from RD Sharma Class 12 Solutions Chapter 3 VSAQ.

5. Where is inverse trigonometry used in real life?

nverse trigonometry is a concept through which we can determine the third side of a triangle if two are known. This has its applications across engineering physics, geometry, and navigation on land and sea. To get more information about Inverse Trigonometry, you can refer to RD Sharma Class 12 Solutions Inverse Trigonometric Functions VSAQ.

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