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

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## 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

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

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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