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

These NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-2-ex-2-1/

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

Ex 2.1 Class 12 Question 1.
sin^-1\(\frac { -1 }{ 2 }\)
Solution:

Question 2.
cos^-1\(\left(\frac{\sqrt{3}}{2}\right)\)
Solution:

Question 3.
cosec^-1(2)
Solution:
The principal values branch of cosec^-1 is

Question 4.
tan^-1(-\(\sqrt{3}\))
Solution:
The principal values branch of tan^-1 is

Question 5.
cos^-1\(\frac { – 1 }{ 2 }\)
Solution:
The principal values branch of cos^-1 is [0, π]

Question 6.
tan^-1(-1)
Solution:
The principal values branch of tan^-1 is

Question 7.
sec^-1\(\left(\frac{2}{\sqrt{3}}\right)\)
Solution:
The principal values branch of sec^-1 is

Question 8.
cot^-1(\(\sqrt{3}\))
Solution:
The principal values branch of cot^-1 is [0, π]
cot\(\frac { π }{ 6 }\) = \(\sqrt{3}\), \(\frac { π }{ 6 }\) ∈ [0, π]
∴ Priciapal value of cot^-1\(\sqrt{3}\) is \(\frac { π }{ 6 }\)

Question 9.
cos^-1\(\left(\frac{-1}{\sqrt{2}}\right)\)
Solution:
The principal values branch of cos^-1 is [0, π]

Question 10.
cosec^-1(\(\sqrt{-2}\))
Solution:
The principal values branch of cosec^-1 is

Question 11.
tan^-1(1) + cos^-1(\(\frac { – 1 }{ 2 }\)) + sin^-1(\(\frac { – 1 }{ 2 }\))
Solution:

Question 12.
cos^-1(\(\frac { 1 }{ 2 }\)) + 2 sin^-1(\(\frac { 1 }{ 2 }\))
Solution:

Question 13.
If sin^-1 x = y, then
a. 0 ≤ y ≤ π
b. – \(\frac { π }{ 2 }\) ≤ y ≤ \(\frac { π }{ 2 }\)
c. 0 < y < π
d. – \(\frac { π }{ 2 }\) < y < \(\frac { π }{ 2 }\)
Solution:
b. – \(\frac { π }{ 2 }\) ≤ y ≤ \(\frac { π }{ 2 }\)
The principal values branch of sin^-1 is
[\(\frac { – π }{ 2 }\), \(\frac { π }{ 2 }\)] . i.e., \(\frac { – π }{ 2 }\) ≤ y ≤ \(\frac { π }{ 2 }\)

Question 14.
tan^-1\(\sqrt{3}\) – sec^-1 ( – 2) is equal to
a. π
b . \(\frac { – π }{ 3 }\)
c. \(\frac { π }{ 3 }\)
d. \(\frac { 2π }{ 3 }\)
Solution:
b . \(\frac { – π }{ 3 }\)
tan^-1\(\sqrt{3}\) – sec^-1 ( – 2) = \(\frac { π }{ 3 }\) – \(\frac { 2π }{ 3 }\) = \(\frac { – π }{ 3 }\)