NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1
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Textbook | NCERT |
Board | CBSE |
Category | NCERT Solutions |
Class | Class 12 |
Subject | Maths |
Chapter | Chapter 2 |
Exercise | Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.1 |
Number of Questions Solved | 14 |

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1
NCERT TEXTBOOK EXERCISES
Ex 2.1 Class 12 Maths Question 1-10.
Find the principal values of the following:
(1) $\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$
(2) $\cos ^{ -1 }{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }$
(3) csc−1(2)
(4) tan−1(–√3)
(5) cos−1(−12)
(6) tan−1(−1)
(7) sec−1(√2/3)
(8)cot−1(√3)
(9) cos−1(−√1/2)
(10) csc−1(−√2)
Solution:
(1) Let $\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$ = y
∴ $sin\quad y=-\frac { 1 }{ 2 } =-sin\frac { \pi }{ 6 } =sin\left( -\frac { \pi }{ 6 } \right)$
the range of principal value of sin-1 is




Ex 2.1 Class 12 Maths Question 11-12.
Find the principal values of the following:
(11) $\tan ^{ -1 }{ (1) } +\cos ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } +\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$
(12) $\cos ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) } +2\sin ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) }$
Solution:
(11) $\tan ^{ -1 }{ (1) } +\cos ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } +\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$
Now tan-1 (1) = π/4
∴the range of principal value branch of

Ex 2.1 Class 12 Maths Question 13.
If sin-1 x = y, then
(a) 0 ≤ y ≤ π
(b) $-\frac { \pi }{ 2 } \le y\le \frac { \pi }{ 2 }$
(c) 0 < y < π
(d) $-\frac { \pi }{ 2 } <y<\frac { \pi }{ 2 }$
Solution:
The range of principal value of sin is [−π/2, π/2]
∴ if sin-1 x = y then
$-\frac { \pi }{ 2 } \le y\le \frac { \pi }{ 2 }$
Option (b) is correct
Ex 2.1 Class 12 Maths Question 14.
tan−1√3−sec−1(−2) is equal to
(a) π
(b) −π/3
(c) π/3
(d) 2π/3
Solution:
$\tan ^{ -1 }{ \sqrt { 3 } =\frac { \pi }{ 3 } ,\sec ^{ -1 }{ (-2) } } =\pi -\frac { \pi }{ 3 } =\frac { 2\pi }{ 3 }$
∴ Principal values of sec-1 is [0,π] – {π/2}
$\tan ^{ -1 }{ \sqrt { 3 } – } \sec ^{ -1 }{ (-2) } =\frac { \pi }{ 3 } -\frac { 2\pi }{ 3 } =-\frac { \pi }{ 3 }$
Option (b) is correct
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.1 PDF
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