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# NCERT Solutions (Exercise 2.1) - Inverse Trigonometric Functions Notes | EduRev

## JEE : NCERT Solutions (Exercise 2.1) - Inverse Trigonometric Functions Notes | EduRev

The document NCERT Solutions (Exercise 2.1) - Inverse Trigonometric Functions Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Q.1. Find the principal value of sin-1 (-1 / 2).
Ans. Let sin-1 (-1 / 2) = y, Then
sin y = -1 / 2 = -sin (Ï€ / 6) = sin (-Ï€ / 6)
We know that the range of the principal value branch of sinâˆ’1 is [ -Ï€ / 2, Ï€ / 2] and sin (- Ï€ / 6) = -1 / 2
Therefore, principal value of sin-1 (-1 / 2) is -Ï€ / 6.

Q.2. Find the principal value of cos-1 (âˆš3 / 2).
Ans. Let cos-1 (âˆš3 / 2) = y, then
cos y = âˆš3 / 2 = cos(Ï€ / 6)
We know that the range of the principal value branch of cos-1 is [0, Ï€] and cos (Ï€ / 6) = âˆš3 / 2
Therefore, principal value of cos-1 (âˆš3 / 2) is Ï€ / 6.

Q.3. Find the principal value of cosecâˆ’1 (2).
Ans. Let cosecâˆ’1 (2) = y. Then, cosec y = 2 = cosec  (Ï€ / 6)
We know that the range of the principal value branch of cosec-1 is [ -Ï€ / 2, Ï€ / 2] - {0} and cosec (Ï€ / 6) = 2.
Therefore, principal value of cos-1 (2) is Ï€ / 6.

Q.4. Find the principal value of tan-1 (âˆš3).
Ans. Let tan-1 (âˆš3) = y, then tan y = -âˆš3 = - tan Ï€ / 3 = tan (-Ï€ / 3)
We know that the range of the principal value branch of tan-1 is [ -Ï€ / 2, Ï€ / 2] and tan (-Ï€ / 3) = -âˆš3
Therefore, principal value of tan-1 (âˆš3) is -Ï€ / 3.

Q.5. Find the principal value of cos-1 (-1 / 2).
Ans. Let cos-1 (-1 / 2) = y, then cos y = -1 / 2 = -cos Ï€ / 3 = cos (Ï€ - Ï€ / 3) = cos (2Ï€ / 3)
We know that the range of the principal value branch of cos-1 is [0, Ï€] and cos(2Ï€ / 3) = -1 / 2
Therefore, principal value of cos-s s--s is sÏ€ / s

Q.6. Find the principal value of tan-1(-1).
Ans. Let tan-1(-1) = y. Then, tan y = -1 = -tan (Ï€ / 4) = tan (-Ï€ / 4)
We know that the range of the principal value branch of tanâˆ’1 is [ -Ï€ / 2, Ï€ / 2] and sin (- Ï€ / 4) = -1
Therefore, principal value of tan-1(-1) is -Ï€ / 4.

Q.7. Find the principal value of sec-1(2 / âˆš3).
Ans. Let sec-1(2 / âˆš3) = y, then sec y = (2 / âˆš3) = sec(Ï€ / 6)
We know that the range of the principal value branch of sec-1 is [0, Ï€] - (Ï€ / 2) and sec ( Ï€ / 6) = 2 / âˆš3
Therefore, principal value of sec-1(2 / âˆš3) is Ï€ / 6.

Q.8. Find the principal value of cot-1âˆš3.
Ans. Let cot-1âˆš3 = y, then cot y = âˆš3 = cot(Ï€ / 6)
We know that the range of the principal value branch of cot-1 is [0, Ï€] and cot ( Ï€ / 6) = âˆš3
Therefore, principal value of cot-1âˆš3 is Ï€ / 6.

Q.9. Find the principal value of cos-1(-1 / âˆš2).
Ans. Let cos-1(-1 / âˆš2) = y, then
cos y = -1 / âˆš2 = -cos(Ï€ / 4) = cos(Ï€ - Ï€ / 4) = cos (3Ï€ / 4)
We know that the range of the principal value branch of cos-1 is [0, Ï€] and cos ( 3Ï€ / 4) = -1 / âˆš2
Therefore, principal value of cos-1(-1 / âˆš2) is 3Ï€ / 4.

Q.10. Find the principal value of cosec-1(-âˆš2).
Ans. Let cosec-1(-âˆš2) = y, then
cosec y = -âˆš2 = -cosec(Ï€ / 4) = cosec(- Ï€ / 4)
We know that the range of the principal value branch of cosec-1 is [-Ï€ / 2, Ï€ / 2] - {0} and cosec ( -Ï€ / 4) = -âˆš2
Therefore, principal value of cosec-1(-âˆš2) is -Ï€ / 4.

Q.11. Find the value of tan-1(1) + cos-1(-1 / 2) + sin-1(-1 / 2).
Ans. Let tan-1(1) = x, then
tan x = 1 = tan Ï€ / 4
We know that the range of the principal value branch of tan-1 is [-Ï€ / 2, Ï€ / 2]
âˆ´ tan-1(1) = Ï€ / 4
Let cos-1(-1 / 2) = y, then
cos y = -1 / 2 = -cos Ï€ / 3 = cos (Ï€ - Ï€ / 3) = cos(2Ï€ / 3)
We know that the range of the principal value branch of cos-1 is [0, Ï€]
âˆ´ cos-1 (-1 / 2) = 2Ï€ / 3
Let sin-1(-1 / 2) = z, then
sin z = -1 / 2 = -sin Ï€ / 6 = sin(-Ï€ / 6)
We know that the range of the principal value branch of sin-1 is [-Ï€ / 2, Ï€ / 2]
âˆ´ sin-1 = (-1 / 2) = -Ï€ / 6
Now,
tan-1(1) + cos-1(-1 / 2) + sin-1(-1 / 2)
= Ï€ / 4 + 2Ï€ / 3 - Ï€ / 6
= 3 x Ï€ + 4 x (2Ï€) - 2(Ï€) / 12
= 3Ï€ + 8Ï€ - 2Ï€ / 12
= 9Ï€ / 12
= 9Ï€ / 12
= 3Ï€ / 4

Q.12.  Find the value of cos-1(1 / 2) + 2sin-1(1 / 2)
Ans. Let cos-1(1 / 2) = x, then cos x = 1 / 2 = cos Ï€ / 3
We know that the range of the principal value branch of cos-1 is [o, Ï€]
âˆ´ cos-1(1 / 2) =  Ï€ / 3
Let sin-1(1 / 2) = y, then sin y = 1 / 2 = sin Ï€ / 6
We know that the range of the principal value branch of sin-1 is [-Ï€ / 2, Ï€ / 2]
âˆ´ sin-1(1 / 2) = Ï€ / 6
cos-1(1 /2) + 2 sin-1(1 / 2) = Ï€ / 3 + 2 x Ï€ / 6 = Ï€ / 3 + Ï€ / 3 = 2Ï€ / 3.

Q.13. If sin-1 x = y, then
(a) 0 â‰¤ y â‰¤ Ï€
(b) -Ï€ / 2 â‰¤ y â‰¤ Ï€ / 2
(c) 0 < y < Ï€
(d) - Ï€ / 2 < y < Ï€ / 2
Ans. (b)
Solution. It is given than sin-1 x = y
We know that the range of the principal value branch of sin-1 is [-Ï€ / 2, Ï€ / 2]
Therefore, -Ï€ / 2 â‰¤ y â‰¤ Ï€ / 2.

Q.14. tan-1âˆš3 - sec-1(-2) is equal to
(a) Ï€
(b) -Ï€ / 3
(c) Ï€ / 3
(d) 2Ï€ / 3
Ans. (b)
Solution. tan-1âˆš3 = x, then
tan x = âˆš3 = tan Ï€ / 3
We know that the range of the principal value branch of tan-1 is [-Ï€ / 2, Ï€ / 2]
âˆ´ tan-1 = Ï€ / 3
Let sec-1(-2) = y, then
sec y = -2 = -sec Ï€ / 3 = sec(Ï€ - Ï€ / 3) = sec(2Ï€ / 3)
We know that the range of the principal value branch of sec-1 is [0, Ï€] - {Ï€ / 2}
âˆ´ sec-1 (-2) = 2Ï€ / 3
Now,
tan-1 âˆš3 - sec-1 (-2) = Ï€ / 3 - 2Ï€ / 3 = -Ï€ / 3

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## Mathematics (Maths) Class 12

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