In this article, we share MP Board Class 12th Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 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 Miscellaneous Exercise

Question 1.

cos^{-1}(cos\(\frac { 13π }{ 6 }\))

Solution:

The principal value brach of cos^{-1} is [0, p]

Question 2.

tan^{-1}(tan\(\frac { 7π }{ 6 }\))

Solution:

The principal value brach of tan^{-1} is (- \(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))

Question 3.

2sin^{-1}\(\frac { 3 }{ 5 }\) = tan^{-1}\(\frac { 24 }{ 7 }\)

Solution:

Question 4.

sin^{-1}\(\frac { 8 }{ 17 }\) + sin^{-1}\(\frac { 3 }{ 5 }\) = tan^{-1}\(\frac { 77 }{ 36 }\)

Solution:

Question 5.

cos^{-1}\(\frac { 4 }{ 5 }\) + cos^{-1}\(\frac { 12 }{ 13 }\) = cos^{-1}\(\frac { 33 }{ 65 }\)

Solution:

Question 6.

cos^{-1}\(\frac { 12 }{ 13 }\) + sin^{-1}\(\frac { 3 }{ 5 }\) = sin^{-1}\(\frac { 56 }{ 65 }\)

Solution:

Question 7.

tan^{-1}\(\frac { 63 }{ 16 }\) = sin^{-1}\(\frac { 5 }{ 13 }\) + cos^{-1}\(\frac { 3 }{ 5 }\)

Solution:

Question 8.

tan^{-1}\(\frac { 1 }{ 5 }\) + tan^{-1}\(\frac { 1 }{ 7 }\) + tan^{-1}\(\frac { 1 }{ 3 }\) + tan^{-1}\(\frac { 1 }{ 8 }\) = \(\frac { π }{ 4 }\)

Solution:

Question 9.

tan^{-1}\(\sqrt{x}\) = \(\frac { 1 }{ 2 }\)cos^{-1}\(\left(\frac{1-x}{1+x}\right), x \in[0,1]\)

Solution:

Question 10.

cot^{-1}\(\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)\) = \(\frac { x }{ 2 }\) x ∈ (0, \(\frac { π }{ 4 }\))

Solution:

Question 11.

tan^{-1}\(\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\) = \(\frac { π }{ 4 }\) – \(\frac { 1 }{ 2 }\) cos^{-1} x, – \(\frac{1}{\sqrt{2}} \leq x \leq 1\)

Solution:

Question 12.

\(\frac { 9p }{ 8 }\) – \(\frac { 9 }{ 4 }\)sin^{-1}\(\frac { 1 }{ 3 }\) = \(\frac { 9 }{ 4 }\)sin^{-1}\(\frac{2 \sqrt{2}}{3}\)

Solution:

Question 13.

2tan^{-1}(cosx) = tan^{-1}(2cosecx)

Solution:

Question 14.

tan^{-1}\(\frac{1-x}{1+x}\) = \(\frac { 1 }{ 2 }\) tan^{-1}x, (x > 0)

Solution:

Question 15.

sin(tan^{-1}x), |x| < 1 is equal to

a. \(\frac{x}{\sqrt{1-x^{2}}}\)

b. \(\frac{1}{\sqrt{1-x^{2}}}\)

c. \(\frac{1}{\sqrt{1+x^{2}}}\)

d. \(\frac{x}{\sqrt{1+x^{2}}}\)

Solution:

d. \(\frac{x}{\sqrt{1+x^{2}}}\)

Question 16.

sin^{-1}(1 – x) – 2sin^{-1}x = \(\frac { π }{ 2 }\), then x is equal to

a. 0, \(\frac { 1 }{ 2 }\)

b. 1, \(\frac { 1 }{ 2 }\)

c. 0

d. \(\frac { 1 }{ 2 }\)

Solution:

c. 0

But x ≠ \(\frac { 1 }{ 2 }\), since sin^{-1}(1 – \(\frac { 1 }{ 2 }\)) ≠ \(\frac { π }{ 2 }\) + 2 sin^{-1}\(\frac { 1 }{ 2 }\)

∴ x = 0.

Question 17.

tan^{-1} \(\frac { x }{ y }\) – tan^{-1} \(\frac{x-y}{x+y}\)

Solution:

a. \(\frac { π }{ 2 }\)

b. \(\frac { π }{ 3 }\)

c. \(\frac { π }{ 4 }\)

d. \(\frac { -3π }{ 4 }\)

Solution:

c. \(\frac { π }{ 4 }\)