Question

NCERT Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions has solutions of important concepts, intext questions, and exercise questions and back of the chapter questions solutions. This NCERT solution is prepared by an expert on the subject matter. Our NCERT Solutions Class 12 covered all the important topics in detail.

• Trigonometric Functions – the trigonometric functions are also defined as circular functions which can be used to derive the functions of any given angle of the triangle. This implies that the relationship between the sides of a triangle with the angle of the triangle is calculated by these trigonometric functions. All the basic functions of trigonometric are sine, cosine, tangent, cotangent, secant, and cosecant functions. There are various trigonometric functions and formulas which can be used to identify the relation of the functions with the help of angles of the triangle. It can also be represented with a table of ratios of some particular degrees. We can solve various trigonometric questions with the help of this table.
• Inverse Trigonometric Functions – the inverse trigonometric function is defined as the inverse of all basic trigonometric functions. The inverse trigonometric functions are also termed the arcus functions, antitrigonometric functions, or cyclometric functions. The main use of these trigonometric functions is to get the angle of any ratio. There are various uses of inverse trigonometric functions in the fields of navigation, engineering, physics, and geometry for various other purposes. There are six types of inverse trigonometric functions Arcsine, arccosine, arctangent, arcsecant, and arccosecant.
• Basic Concepts – the basic inverse trigonometric functions are defined by the sine, cosine, tangent, cotangent, secant, and cosecant functions. The basic trigonometric functions can be defined as cyclometric functions, antitrigonometric functions, arcus functions, etc.

NCERT Solutions Class 12 Maths will help students gain a complete idea of all the important concepts in detail.

Answer 1

NCERT Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions

1. Find the principal value of sin^-1\(\left(-\frac12\right)\) 

Answer:

Let  sin^-1\(\left(-\frac12\right)\) = y, then sin y = \(-\frac12\) = - sin\(\left(\frac{\pi}{6}\right)\) = sin \(\left(-\frac{\pi}{6}\right)\) 

We know that the range of the principal value branch of sin^-1 is \(\left[-\frac{\pi}{2},\frac{\pi}2\right]\) and  sin \(\left(-\frac{\pi}{6}\right)\) = \(-\frac12\) 

Therefore, the principal value of sin^-1 \(\left(-\frac12\right)\) is \(-\frac{\pi}{6}\).

2. Find the principal value of cos^-1\(\left(\frac{\sqrt3}2\right)\)

Answer:

Let  cos^-1\(\left(\frac{\sqrt3}2\right)\) = y, then cos y = \(\frac{\sqrt3}2\) = cos \(\left(\frac{\pi}{6}\right)\)  

We know that the range of the principal value branch of cos^-1 is \(\left[0,\pi\right]\) and  cos \(\left(\frac{\pi}{6}\right)\) = \(\frac{\sqrt3}2\) 

Therefore, the principal value of  cos^-1\(\left(\frac{\sqrt3}2\right)\) is \(\frac{\pi}{6}\).

3. Find the principal value of cosec^-1 (2).

Answer:

Let  cosec^-1 (2) = y, then cosec y = 2 = cosec \(\left(\frac{\pi}{6}\right)\)  

We know that the range of the principal value branch of cosec^-1 is \(\left[-\frac{\pi}{2},\frac{\pi}2\right]\) and  cosec \(\left(\frac{\pi}{6}\right)\) = 2

Therefore, the principal value of cosec^-1 (2)   is \(\frac{\pi}{6}\).

4. Find the principal value of tan^-1 (-√3).

Answer:

Let  tan^-1 (-√3) = y, then tan y = -√3 = - tan \(\frac{\pi}{3}\) = tan \(\left(-\frac{\pi}{3}\right)\) 

We know that the range of the principal value branch of tan^-1 \(\left[-\frac{\pi}{2},\frac{\pi}2\right]\) and tan \(\left(-\frac{\pi}{3}\right)\) = -√3

Therefore, the principal value of tan^-1 (-√3) = \(-\frac{\pi}{3}\).

5. Find the principal value of cos^-1 \(\left(-\frac12\right)\). 

Answer:

Let  cos^-1\(\left(-\frac12\right)\) = y, then cos y = \(-\frac12\) =  - cos \(\frac{\pi}{3}\) = cos \(\left(\pi-\frac{\pi}{3}\right)\) = cos \(\left(\frac{2\pi}{3}\right)\) 

We know that the range of the principal value branch of cos^-1 is \(\left[0,\pi\right]\) and  cos \(\left(\frac{2\pi}{3}\right)\) = \(-\frac12\)

Therefore, the principal value of  cos^-1 \(\left(-\frac12\right)\) is \(\frac{2\pi}{3}\).

6. Find the principal value of tan^-1 (-1).

Answer:

Let  tan^-1 (-1) = y, then tan y = -1 = - tan \(\frac{\pi}{4}\) = tan \(\left(-\frac{\pi}{4}\right)\) 

We know that the range of the principal value branch of tan^-1 \(\left[-\frac{\pi}{2},\frac{\pi}2\right]\) and tan \(\left(-\frac{\pi}{4}\right)\) = -1

Therefore, the principal value of tan^-1 (-1) is \(-\frac{\pi}{4}\).

7. Find the principal value of sec^-1 \(\left(\frac2{\sqrt3}\right)\).

Answer:

Let sec^-1\(\left(\frac2{\sqrt3}\right)\) = y, then sec y = \(\frac{\sqrt3}2\) = cos \(\left(\frac{\pi}{6}\right)\)

We know that the range of the principal value branch of sec^-1 is \(\left[0,\pi\right]\) - \(\left(\frac{\pi}{2}\right)\) and  sec \(\left(\frac{\pi}{6}\right)\) = \(\frac2{\sqrt3}\)

Therefore, the principal value of sec^-1\(\left(\frac2{\sqrt3}\right)\) is \(\frac{\pi}{6}\).

8. Find the principal value of cot^-1 √3.

Answer:

Let  cot^-1 (√3) = y, then cot y = √3 = cot \(\left(\frac{\pi}{6}\right)\).

We know that the range of the principal value branch of cot^-1 is \(\left[0,\pi\right]\) and  cot \(\left(\frac{\pi}{6}\right)\) = √3

Therefore, the principal value of cot^-1 √3 is \(\frac{\pi}{6}\).

9. Find the principal value of cos^-1 \(\left(-\frac1{\sqrt2}\right)\). 

Answer:

Let  cos^-1\(\left(-\frac1{\sqrt2}\right)\) = y, then cos y = \(-\frac1{\sqrt2}\) =  - cos \(\frac{\pi}{4}\) = cos \(\left(\pi-\frac{\pi}{4}\right)\) = cos \(\left(\frac{3\pi}{4}\right)\).

We know that the range of the principal value branch of cos^-1 is \(\left[0,\pi\right]\) and  cos \(\left(\frac{3\pi}{4}\right)\) = \(-\frac1{\sqrt2}\).

Therefore, the principal value of  cos^-1 \(\left(-\frac1{\sqrt2}\right)\) is \(\frac{3\pi}{4}\).

10. Find the principal value of cosec^-1 (-√2).

Answer:

Let  cosec^-1 (-√2) = y, then cosec y = -√2 = -cosec \(\left(\frac{\pi}{4}\right)\) = cosec \(\left(-\frac{\pi}{4}\right)\)

We know that the range of the principal value branch of cosec^-1 is \(\left[-\frac{\pi}{2},\frac{\pi}2\right]\) - {0} and  cosec \(\left(-\frac{\pi}{4}\right)\) = √2

Therefore, the principal value of cosec^-1 (-√2) is \(-\frac{\pi}{4}\).

Answer 2

11. Find the value of tan^-1 + cos^-1 \(\left(-\frac1{2}\right)\) + sin^-1 \(\left(-\frac1{2}\right)\).

Answer:

Let tan^-1 (1) = x, then tan x = 1 = tan\(\frac{\pi}{4}\) 

We know that the range of the principal value branch of tan^-1 \(\left(-\frac{\pi}{2},\frac{\pi}2\right)\).

\(\therefore\) tan^-1 (1) = \(\frac{\pi}4\) 

Let  cos^-1 \(\left(-\frac1{2}\right)\) = y, then

We know that the range of the principal value branch of cos^-1 is \(\left[0,\pi\right]\).

\(\therefore \) cos^-1 \(\left(-\frac1{2}\right)\) = \(\frac{2\pi}{3}\) 

Let sin^-1 = \(\left(-\frac1{2}\right)\) = z, then 

We know that the range of the principal value branch of sin^-1 is \(\left[-\frac{\pi}{2},\frac{\pi}2\right]\).

 \(\therefore \) sin^-1 \(\left(-\frac1{2}\right)\) = \(-\frac{\pi}{6}\)  

Now,

tan^-1 + cos^-1 \(\left(-\frac1{2}\right)\) + sin^-1 \(\left(-\frac1{2}\right)\) 

12.  Find the value of cos^-1 \(\left(\frac1{2}\right)\) + 2sin^-1 \(\left(\frac1{2}\right)\).

Answer:

Let cos^-1 \(\left(\frac1{2}\right)\) = x, then

cos x = \(\frac12\) = cos\(\frac{\pi}3\) 

We know that the range of the principal value branch of cos^-1 is \(\left[0,\pi\right]\).

\(\therefore \) cos^-1 \(\left(\frac1{2}\right)\) = \(\frac{\pi}3\) 

Let sin^-1 = \(\left(-\frac1{2}\right)\) = y, then 

sin y = \(\frac12\) = cos \(\frac{\pi}3\)

We know that the range of the principal value branch of sin^-1 is \(\left[-\frac{\pi}{2},\frac{\pi}2\right]\).

 \(\therefore \) sin^-1 \(\left(\frac1{2}\right)\) = \(\frac{\pi}{6}\)  

Now,

13. If sin^-1 x = y, then

(A) \(0\le y\le \pi\)

(B) \(-\frac{\pi}2 \le y \le \frac{\pi}{2}\) 

(C) \(0 < y <\pi\) 

(D) \(-\frac{\pi}2 < y< \frac{\pi}2\) 

Answer:

It is given that sin^-1 x = y

We know that the range of the principal value branch of sin^-1 is \(\left[-\frac{\pi}{2},\frac{\pi}2\right]\).

Therefore, \(-\frac{\pi}2 \le y \le \frac{\pi}{2}\).

Hence, the option (B) is correct.

14. tan^-1 √3 - sec^-1 (-2) is equal to 

(A) \(\pi\)

(B) \(-\frac{\pi}{3}\)

(C) \(\frac{\pi}{3}\)

(D) \(\frac{2\pi}{3}\)

Answer:

Let tan^-1 √3 = x, then

tan x = √3 = tan \(\frac{\pi}3\) 

 We know that the range of the principal value branch of tan^-1 is \(\left[-\frac{\pi}{2},\frac{\pi}2\right]\).

 \(\therefore \) tan^-1 √3 = \(\frac{\pi}3\)

Let sec^-1 (-2) = y, then

We know that the range of the principal value branch of sec^-1 is \(\left[0,\pi\right]\) - \(\left[\frac{\pi}3\right]\)

\(\therefore \) sec^-1 (-2) = \(\frac{2\pi}3\) 

Now,

Hence, the option (B) is correct.

15. Prove that 3sin⁻¹ x = sin⁻¹ (3x − 4x³); x ∈ \(\left[-\frac12,\frac12\right]\)

Answer:

Let sin^-1 x = \(\theta\) , then x =  sin \(\theta\). We have,

RHS = sin^-1(3x - 4x³) = sin^-1 (3 sin \(\theta\) - 4sin³\(\theta\))

= sin^-1 (sin 3\(\theta\)) = 3\(\theta\) = 3sin^-1x = LHS

16. Prove that 3cos⁻¹x = cos⁻¹(4x³ − 3x), x ∈ \(\left[\frac12,1\right]\)

Answer:

Let cos^-1 x = θ

x = cos θ

R.H.S= cos^-1 (4x³ – 3cosx)

= cos^-1 (4 cos³θ – 3 cosθ)

= cos^-1 (cos 3θ) [∴ cos 3θ = 4 cos³θ – 3 cos θ]

= 3θ

= 3 cos^-1 x

= L.H.S.

17. Prove that tan^-1 \(\frac2{11}\) + tan^-1 \(\frac7{24}\) = tan^-1 \(\frac12\)

Answer:

18. Prove that 2tan^-1 \(\frac12\) + tan^-1 \(\frac17\) = tan^-1 = \(\frac{31}{17}\) 

Answer:

Answer 3

19. Write the functions \(\text{tan}^{-1}\frac{\sqrt{1+x^2}-1}{x}\) , \(x \ne0\), in the simplest form.

Answer:

Given function \(\text{tan}^{-1}\frac{\sqrt{1+x^2}-1}{x}\) 

Let x = tan θ

20.  Write the functions \(\text{tan}^{-1}\frac1{\sqrt{x^2 -1}}\) , \(|x| >1\), in the simplest form.

Answer:

Given function \(\text{tan}^{-1}\frac1{\sqrt{x^2 -1}}\) 

Let x = cosec θ

21. Write the functions \(\text{tan}^{-1}\left(\sqrt{\frac{1-cos\,x}{1+ cos\,x}}\right)\) , \(x >\pi\), in the simplest form.

Answer:

The given function is \(\text{tan}^{-1}\left(\sqrt{\frac{1-cos\,x}{1+ cos\,x}}\right)\) 

Now,

Answer 4

22. Write the function \(\text{tan}^{-1}\left(\frac{cos\,x- sin\,x}{cos\,x + sin\,x}\right)\)\(0< x<\pi\) , in the simplest form.

Answer:

The given function is \(\text{tan}^{-1}\left(\frac{cos\,x- sin\,x}{cos\,x + sin\,x}\right)\)

Now,

23. Write the functions \(\text{tan}^{-1}\frac x{\sqrt{a^2 -x^2}}\), \(|x| <a\), in the simplest form.

Answer:

The given function is \(\text{tan}^{-1}\frac x{\sqrt{a^2 -x^2}}\)

Let x = a sin θ

24. Write the functions \(​​\text{tan}^{-1}\left(\frac{3a^2x - x^3}{a^3- 3ax^2}\right), a>0; \frac{-a}{\sqrt3}\le x \frac a{\sqrt3},\) in the simplest form.

Answer:

The given function is \(\text{tan}^{-1}\left(\frac{3a^2x - x^3}{a^3- 3ax^2}\right)\)

Let x = a tan θ

Answer 5

25. Find the value of \(\text{tan}^{-1} [2\,cos (2\,sin^{-1}\frac12)]\) 

Answer:

The given function is \(\text{tan}^{-1} [2\,cos (2\,sin^{-1}\frac12)]\) 

26. Find the value of cot (tan^-1 a + cot^-1 a).

Answer:

The given function is cot (tan^-1 a + cot^-1 a)

\(\therefore \) cot (tan^-1 a + cot^-1 a) = cot \(\left(\frac{\pi}{2}\right)\) = 0    [as tan^-1 x + cot^-1 x = \(\frac{\pi}{2}\)]

27. Find the value of \(tan\frac12\left[[sin^{-1} \cfrac{2x}{1+x^2 }+ cos^{-1} \frac{1-y^2}{1+y^2}\right]\), \(|x| <1, y>0 \,\text{and}\,xy <1.\) 

Answer:

The given function is \(tan\frac12\left[[sin^{-1} \cfrac{2x}{1+x^2 }+ cos^{-1} \frac{1-y^2}{1+y^2}\right]\) 

28. If sin (sin^-1 \(\frac15\) + cos^-1 x) = 1, then find the value of x.

Answer:

Since, sin (sin^-1 \(\frac15\) + cos^-1 x) = 1

29. If \(tan^{-1}\frac{x-1}{x-2} + tan^{-1}\frac{x+1}{x+2} = \frac{\pi}{4}\), then find the value of x.

Answer:

Given that \(tan^{-1}\frac{x-1}{x-2} + tan^{-1}\frac{x+1}{x+2} = \frac{\pi}{4}\)

30. Find the values of \(sin^{-1} \left(sin\frac{2\pi}{3}\right)\).

Answer:

Given that \(sin^{-1} \left(sin\frac{2\pi}{3}\right)\). 

We know that sin^-1 (sin x) =  x if x \(\in \left[-\frac{\pi}{2},\frac{\pi}2\right]\), which is the principal value of branch of sin^-1 x.

31. Find the values of \(\tan^{-1}\left(tan\frac{3\pi}{4}\right)\).

Answer:

Given that \(\tan^{-1}\left(tan\frac{3\pi}{4}\right)\) 

We know that tan^-1 (tan x) =  x if x \(\in \left[-\frac{\pi}{2},\frac{\pi}2\right]\), which is the principal value of branch of tan^-1 x.

32. Find the values of \(tan(sin^{-1}\frac35 + cot^{-1} \frac32)\).

Answer:

Given that \(tan(sin^{-1}\frac35 + cot^{-1} \frac32)\) 

33. \(cos^{-1}\left(cos\frac{7\pi}{6}\right)\) is equal to

(A) \(\frac{7\pi}6\)

(B) \(\frac{5\pi}6\)

(C) \(\frac{\pi}3\)

(D) \(\frac{\pi}6\)

Answer:

Given that \(cos^{-1}\left(cos\frac{7\pi}{6}\right)\) 

We know that cos^-1 (cos x) = x, if x \(\in[0,\pi]\), which is the principal value branch of cos^-1 x.

Hence, the option (B) is correct.

34. \(sin\left(\frac{\pi}{3}- sin^{-1}(-\frac12)\right) \) is equal to

(A) \(\frac12\)

(B) \(\frac13\)

(C) \(\frac14\)

(D) 1

Answer:

Given that \(sin\left(\frac{\pi}{3}- sin^{-1}(-\frac12)\right) \) 

We know that the range of the principle value branch of sin^-1 is\(\left[-\frac{\pi}{2},\frac{\pi}2\right]\).

Hence, the option (D) is correct.

35. tan^-1 √3 - cot^-1 (-√3) is equal to

(A) \(\pi\)

(B) \(- \frac{\pi}{2}\)

(C) 0

(D) 2√3

Answer:

Given that tan^-1 √3 - cot^-1 (-√3)

We know that the range of the principle value branch of tan^-1 is\(\left(-\frac{\pi}{2},\frac{\pi}2\right)\) and cot^-1 is(0, \(\pi\)).

Hence, the option (B) is correct.

Answer 6

36. Find the value of \(cos^{-1}\left(cos\frac{13\pi}{6}\right)\).

Answer:

Given that \(cos^{-1}\left(cos\frac{13\pi}{6}\right)\)

We know that cos^-1 (cos x) = x if x \(\in [0,\pi]\), which is the principal value branch of cos^-1 x.

37. Find the value of \(tan^{-1} \left(tan\frac{7\pi}{6}\right)\).

Answer:

Given that \(tan^{-1} \left(tan\frac{7\pi}{6}\right)\) 

We know that tan^-1 (tan x) = x if x \(\in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), which is the principal value branch of tan^-1 x.

38. Prove that \(2sin^{-1} \frac35 = tan^{-1}\frac{24}{7}\).

Answer:

39. Prove that \(sin^{-1} \frac8{17}+sin^{-1}\frac35= tan^{-1}\frac{77}{36}\).

Answer:

40. Prove that \(cos^{-1} \frac45+ cos^{-1}\frac{12}{13}= cos^{-1}\frac{33}{65}\)

Answer:

41. Prove that \(cos^{-1}\frac{12}{13}+ sin^{-1}\frac35 = sin^{-1}\frac{56}{65}\)

Answer:

42. Prove that \(tan^{-1}\frac{63}{16}= sin^{-1}\frac5{13} + cos^{-1}\frac35\) 

Answer:

43. Prove that \(tan^{-1}\frac15 + tan^{-1} \frac17 + tan^{-1}\frac18 = \frac{\pi}{4}\) 

Answer:

44. Prove that \(tan^{-1}\sqrt x = \frac12 cos^{-1}\left(\frac{1-x}{1+x}\right),x\in[0,1]\)

Answer:

45. Prove that \(cot^{-1}\left(\frac{\sqrt{1+sin\,x}+\sqrt{1-sin\,x}}{\sqrt{1+sin\,x}-\sqrt{1-sin\,x}}\right)= \frac x2, x\in\left(0,\frac{\pi}{4}\right)\) 

Answer:

Answer 7

46. Prove that \(tan^{-1}\left(\frac{\sqrt{1+x} -\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)= \frac{\pi}{4}-\frac12 cos^{-1}x,-\frac1{\sqrt2}\le x\le 1\). 

Answer:

47. Prove that \(\frac{9\pi}{8}- \frac94\,sin^{-1}\frac13= \frac94 \,sin^{-1}\frac{2\sqrt2}{3}\) 

Answer:

48. Solve for x: 2tan^-1(cos x) = tan^-1(2cosec x)

Answer:

Given that 2tan^-1(cos x) = tan^-1(2cosec x)

But sin x \(\ne\) 0 as it does not satisfy the equation

\(\therefore\) cos x -  sin x = 0 

⇒ cos x = sin x 

⇒ tan x = 1

\(\therefore\) \(x = \frac{\pi}4\)

49. Solve for x: \(tan^{-1}\frac{1-x}{1+x} = \frac12tan^{-1}x, (x > 0)\) 

Answer:

Given that: \(tan^{-1}\frac{1-x}{1+x} = \frac12tan^{-1}x, (x > 0)\) 

50. 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) \(\frac{x}{\sqrt{1+x^2}}\)

Answer:

Given that: sin(tan^-1 x)

Hence, the option (D) is correct.

51. sin^-1(1 - x) - 2sin^-1x = \(\frac{\pi}{2}\), then x is equal to

(A) 0, \(\frac12\)

(B) 1, \(\frac12\)

(C) 0

(D) \(\frac12\)

Answer:

Given that: sin^-1(1 - x) - 2sin^-1x = \(\frac{\pi}{2}\)

Let x = sin y

But x \(\ne\) \(\frac12\), as it does not satisfy the given equation.

\(\therefore\) x = 0 is the solution of the given equation.

Hence, the option (C) is correct.

52. \(tan^{-1}\left(\frac xy\right) - tan^{-1}\frac{x-y}{x+y}\) is equal to

(A) \(\frac{\pi}{2}\)

(B) \(\frac{\pi}{3}\)

(C) \(\frac{\pi}4\)

(D) \(-\frac{3\pi}4\)

Answer:

Hence, the option (C) is correct.