NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1.
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 12 |
| Subject | Maths |
| Chapter | Chapter 2 |
| Chapter Name | Inverse Trigonometric Functions |
| Exercise | Ex 2.1 |
| Number of Questions Solved | 14 |
| Category | NCERT Solutions |
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1
You can find the derivative of an Inverse Function Calculator to solve your equations online and learn quickly.
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) \(\tan ^{ -1 }{ \left( -\sqrt { 3 } \right) } \)
(5) \(\cos ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } \)
(6) \(\tan ^{ -1 }{ (-1) } \)
(7) \(\sec ^{ -1 }{ \left( \frac { 2 }{ \sqrt { 3 } } \right) } \)
(8)\(\cot ^{ -1 }{ \left( \sqrt { 3 } \right) } \)
(9) \(\cos ^{ -1 }{ \left( -\frac { 1 }{ \sqrt { 2 } } \right) } \)
(10) \(\csc ^{ -1 }{ \left( -\sqrt { 2 } \right) } \)
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) = \(\frac { \pi }{ 4 } \)
∴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 \(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
∴ 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.
\(\tan ^{ -1 }{ \sqrt { 3 } -\sec ^{ -1 }{ (-2) } } \) is equal to
(a) π
(b) \(-\frac { \pi }{ 3 } \)
(c) \(\frac { \pi }{ 3 } \)
(d) \(\frac { 2\pi }{ 3 } \)
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] – \(\left\{ \frac { \pi }{ 2 } \right\} \)
\(\tan ^{ -1 }{ \sqrt { 3 } – } \sec ^{ -1 }{ (-2) } =\frac { \pi }{ 3 } -\frac { 2\pi }{ 3 } =-\frac { \pi }{ 3 } \)
Option (b) is correct
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