In this article, we share MP Board Class 12th Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.1 Pdf, These solutions are solved by subject experts from the latest MP Board books.

## MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.1

Find the principal values of the following.

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) = \(\frac { π }{ 3 }\) – \(\frac { 2π }{ 3 }\) = \(\frac { – π }{ 3 }\)