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

In this chapter, we provide NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 pdf, free NCERT solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 book pdf download.

TextbookNCERT
ClassClass 12
SubjectMaths
ChapterChapter 2
Chapter NameInverse Trigonometric Functions
ExerciseEx 2.1
CategoryNCERT Solutions

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

Ex 2.1 Class 12 Maths Question 1-10.
Find the principal values of the following:
(1) 
(2) 
(3) 
(4) 
(5) 
(6) 
(7) 
(8)
(9) 
(10) 
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
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Q1.1
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Q1.2
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Q1.3
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Q1.4

Ex 2.1 Class 12 Maths Question 11-12.
Find the principal values of the following:
(11) 
(12) 
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
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Q2.1

Ex 2.1 Class 12 Maths Question 13.
If sin-1 x = y, then
(a) 0 ≤ y ≤ π
(b) 
(c) 0 < y < π
(d) 
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.
 is equal to
(a) π
(b) 
(c) 
(d) 
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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