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

Mathematics (Maths) Class 12

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 (π / 6) = sin (-π / 6)
We know that the range of the principal value branch of sin−1 is [ -π / 2, π / 2] and sin (- π / 6) = -1 / 2
Therefore, principal value of sin-1 (-1 / 2) is -π / 6.

Q.2. Find the principal value of cos-1 (√3 / 2).
Ans. Let cos-1 (√3 / 2) = y, then
cos y = √3 / 2 = cos(π / 6)
We know that the range of the principal value branch of cos-1 is [0, π] and cos (π / 6) = √3 / 2
Therefore, principal value of cos-1 (√3 / 2) is π / 6.

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

Q.4. Find the principal value of tan-1 (√3).
Ans. Let tan-1 (√3) = y, then tan y = -√3 = - tan π / 3 = tan (-π / 3)
We know that the range of the principal value branch of tan-1 is [ -π / 2, π / 2] and tan (-π / 3) = -√3
Therefore, principal value of tan-1 (√3) is -π / 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 π / 3 = cos (π - π / 3) = cos (2π / 3)
We know that the range of the principal value branch of cos-1 is [0, π] and cos(2π / 3) = -1 / 2
Therefore, principal value of cos-s s--s is sπ / s

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

Q.7. Find the principal value of sec-1(2 / √3).
Ans. Let sec-1(2 / √3) = y, then sec y = (2 / √3) = sec(π / 6)
We know that the range of the principal value branch of sec-1 is [0, π] - (π / 2) and sec ( π / 6) = 2 / √3
Therefore, principal value of sec-1(2 / √3) is π / 6.

Q.8. Find the principal value of cot-1√3.
Ans. Let cot-1√3 = y, then cot y = √3 = cot(π / 6)
We know that the range of the principal value branch of cot-1 is [0, π] and cot ( π / 6) = √3
Therefore, principal value of cot-1√3 is π / 6.

Q.9. Find the principal value of cos-1(-1 / √2).
Ans. Let cos-1(-1 / √2) = y, then
cos y = -1 / √2 = -cos(π / 4) = cos(π - π / 4) = cos (3π / 4)
We know that the range of the principal value branch of cos-1 is [0, π] and cos ( 3π / 4) = -1 / √2
Therefore, principal value of cos-1(-1 / √2) is 3π / 4.

Q.10. Find the principal value of cosec-1(-√2).
Ans. Let cosec-1(-√2) = y, then
cosec y = -√2 = -cosec(π / 4) = cosec(- π / 4)
We know that the range of the principal value branch of cosec-1 is [-π / 2, π / 2] - {0} and cosec ( -π / 4) = -√2
Therefore, principal value of cosec-1(-√2) is -π / 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 π / 4
We know that the range of the principal value branch of tan-1 is [-π / 2, π / 2]
∴ tan-1(1) = π / 4
Let cos-1(-1 / 2) = y, then
cos y = -1 / 2 = -cos π / 3 = cos (π - π / 3) = cos(2π / 3)
We know that the range of the principal value branch of cos-1 is [0, π]
∴ cos-1 (-1 / 2) = 2π / 3
Let sin-1(-1 / 2) = z, then
sin z = -1 / 2 = -sin π / 6 = sin(-π / 6)
We know that the range of the principal value branch of sin-1 is [-π / 2, π / 2]
∴ sin-1 = (-1 / 2) = -π / 6
Now,
tan-1(1) + cos-1(-1 / 2) + sin-1(-1 / 2)
= π / 4 + 2π / 3 - π / 6
= 3 x π + 4 x (2π) - 2(π) / 12
= 3π + 8π - 2π / 12
= 9π / 12
= 9π / 12
= 3π / 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 π / 3
We know that the range of the principal value branch of cos-1 is [o, π]
∴ cos-1(1 / 2) =  π / 3
Let sin-1(1 / 2) = y, then sin y = 1 / 2 = sin π / 6
We know that the range of the principal value branch of sin-1 is [-π / 2, π / 2]
∴ sin-1(1 / 2) = π / 6
cos-1(1 /2) + 2 sin-1(1 / 2) = π / 3 + 2 x π / 6 = π / 3 + π / 3 = 2π / 3.

Q.13. If sin-1 x = y, then
(a) 0 ≤ y ≤ π
(b) -π / 2 ≤ y ≤ π / 2 
(c) 0 < y < π
(d) - π / 2 < y < π / 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 [-π / 2, π / 2]
Therefore, -π / 2 ≤ y ≤ π / 2.

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

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