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Maths XII chapter 2 inverse trigonometric functions ex 2 1

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Chapter 2
Mathematics XII
Exercise 2.1
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Exercise 2.1
Question 1: www.tiwariacademy.com
1
Find the principal value of −1 (− )
2
Answer 1:
1
Let −1 (− ) = , Then
2
1


  = − = − ( ) =  (− )
2
6
6
π π
We know that the range of the principal value branch of sin −1 is [- , ] and
2 2
π
1
sin (- ) = 6
2
1
π
Therefore, the principal value of sin-1 (- ) is - .
2
6
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Question 2:
√3
Find the principal value of  −1 ( )
2
Answer 2: www.tiwariacademy.com
√3
Let  −1 ( ) = , then
2

√3
=  ( )
2
6
We know that the range of the principal value branch of cos −1 is [0, π] and
cos  =
π
cos ( ) =
6
√3
2
√3
π
Therefore, the principal value of cos -1 ( ) is .
2
6
1
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Question 3: www.tiwariacademy.com
Find the principal value of cosec−1 (2).
Answer 3:
π
Let cosec−1 (2) = y. Then, cosec y = 2 = cosec ( )
6
π π
We know that the range of the principal value branch of cosec −1 is [- 2 , 2 ] -{0}
π
and cosec ( ) = 2.
6
π
Therefore, the principal value of cosec−1 (2) is .
6
Question 4: www.tiwariacademy.com
Find the principal value of −1 (−√3).
Answer 4:
π
π
Let tan-1 (-√3) = y, then tan y = -√3 = -tan = tan (- )
3
3
π π
We know that the range of the principal value branch of tan−1 is (- , ) and
2 2
π
tan (- ) = -√3
3
π
Therefore, the principal value of tan-1 (-√3) is - .
3
Question 5: www.tiwariacademy.com
1
Find the principal value of  −1 (− ).
2
Answer 5:
1
1
π
π
2π
Let cos -1 (- ) = y, then cos y = - = -cos = cos (π- ) = cos ( )
2
2
3
3
3
We know that the range of the principal value branch of cos −1 is [0, π] and
2π
1
cos ( ) = 3
2
1
Therefore, the principal value of cos -1 (- ) is
2
2π
3
.
2
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Question 6: www.tiwariacademy.com
Find the principal value of tan−1 (−1).
Answer 6:
π
π
Let tan−1 (−1) = y. Then, tan y = -1 = -tan ( ) = tan (- )
4
4
π π
We know that the range of the principal value branch of tan −1 is (- , ) and
2 2
π
tan (- ) = -1
4
π
Therefore, the principal value of tan−1 (−1) is - .
4
Question 7: www.tiwariacademy.com
2
Find the principal value of  −1 ( ).
3
√
Answer 7:
2
2
√
√
Let  −1 ( ) = , then sec  =
3

=  ( )
3
6

We know that the range of the principal value branch of sec −1 is [0, ] − { }
2

and  ( ) =
6
2
√3
.
2

Therefore, the principal value of  −1 ( ) is .
3
6
√
Question 8: www.tiwariacademy.com
Find the principal value of  −1 √3.
Answer 8:
π
Let cot -1 √3 = y, then cot y = √3 = cot ( ).
6
We know that the range of the principal value branch of cot −1 is (0, π) and
π
cot ( ) = √3.
6
π
Therefore, the principal value of cot -1 √3 is .
6
3
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Question 9: www.tiwariacademy.com
Find the principal value of  −1 (−
1
).
√2
Answer 9:
Let cos -1 (-
cos y = -
1
) = y, then
√2
1
√
π
π
3π
= -cos ( ) = cos (π- ) = cos ( ).
2
4
4
4
We know that the range of the principal value branch of cos −1 is [0, π] and
3π
cos ( ) = 4
1
.
√2
Therefore, the principal value of cos -1 (-
1
) is
√2
3π
4
.
Question 10:
Find the principal value of  −1 (−√2).
Answer 10:
Let  −1 (−√2) = , then


  = −√2 = − ( ) =  (− )
4
4
π π
We know that the range of the principal value branch of cosec−1 is [- , ] -{0}
2 2
π
and cosec (- ) = -√2.
4
π
Therefore, the principal value of cosec -1 (-√2) is - .
4
4
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Question 11: www.tiwariacademy.com
1
1
Find the value of −1 (1) +  −1 (− ) + −1 (− ).
2
2
Answer 11:
Let −1 (1) = , then
tan  = 1 = 

4
 
We know that the range of the principal value branch of tan−1 is (− , ) .
2 2

∴ −1 (1) =
4
1
Let  −1 (− ) = , then
2
1


2
cos  = − = − =  ( − ) =  ( )
2
3
3
3
−1
We know that the range of the principal value branch of cos is [0, ].
1
∴  −1 (− ) =
2
2
3
1
Let −1 (− ) = , then
2
1


sin  = − = − =  (− )
2
6
6
π π
We know that the range of the principal value branch of sin−1 is [- , ].
2 2
1
π
∴ sin-1 (- ) = 2
6
Now,
1
1
tan-1 (1) + cos -1 (- ) + sin-1 (- )
2
2
=
 2  3 + 8 − 2 9 3
+
− =
=
=
4
3
6
12
12
4
5
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Question 12: www.tiwariacademy.com
1
1
Find the value of  −1 ( ) + 2−1 ( )
2
2
Answer 12:
1
1
π
Let cos -1 ( ) = x, then cos x = = cos
2
2
3
We know that the range of the principal value branch of cos−1 is [0, π].
1
∴ cos -1 ( ) =
2
π
3
1
1
π
Let sin-1 (- ) = y, then sin y = = sin
2
2
6
π π
We know that the range of the principal value branch of sin−1 is [- , ].
2 2
∴
1
−1 ( )
2
=

6
Now,
1
1

   2
 −1 ( ) + 2−1 ( ) = + 2 × = + = .
2
2
3
6 3 3
3
Question 13:
If sin−1 x = y, then


2

2

2
2
(A) 0 ≤  ≤ 
(B) − ≤  ≤
(C) 0 <  < 
(D) − <  <
Answer 13: www.tiwariacademy.com
It is given that sin−1 x = y.
π π
We know that the range of the principal value branch of sin−1 is [- , ].
2 2
π
π
2
2
Therefore, - ≤ y ≤ .
Hence, the option (B) is correct.
6
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Question 14: www.tiwariacademy.com
−1 √3 −  −1 (−2) is equal to
(A) π
(C)
(B) −

(D)
3

3
2
3
Answer 14:
Let −1 √3 = , then
tan  = √3 = 

3
π π
We know that the range of the principal value branch of tan−1 is (- , ).
2 2
π
∴ tan-1 √3 =
3
Let sec -1 (-2) = y, then
sec y = -2 = -sec
π
π
2π
= sec (π- ) = sec ( )
3
3
3
π
We know that the range of the principal value branch of sec−1 is [0, π]- { }
2
∴ sec -1 (-2) =
2π
3
Now,
tan-1 √3-sec -1 (-2) =
π 2π
π
=3 3
3
Hence, the option (B) is correct. www.tiwariacademy.com
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