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NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1

Here, Below you all know about NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Question Answer. I know many of you confuse about finding Chapter 2 Inverse Trigonometric Functions Ex 2.1 Of Class 12 NCERT Solutions. So, Read the full post below and get your solutions.

TextbookNCERT
BoardCBSE
CategoryNCERT Solutions
ClassClass 12
SubjectMaths
ChapterChapter 2
ExerciseClass 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.1
Number of Questions Solved14
NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1

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−13−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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