Important Questions for CBSE Class 12 Maths Chapter 2 – Inverse Trigonometric Functions 2023-24

Inverse Trigonometric Functions Class 12 Important Questions for CBSE Maths Chapter 2

Important Questions Class 12 Maths Chapter 2 on Inverse trigonometry are made by Extramarks subject matter experts in accordance with the CBSE syllabus. Maths Class 12 Chapter 2 Important Questions include principal Inverse Trigonometric functions with domain and range, the properties of Inverse Trigonometric functions

These questions come with concise answers and cover all of the necessary and important concepts which will help students to prepare for exams. Students can refer to Class 12 Maths Chapter 2 Important Questions on the Extramarks website.

You can also find CBSE Class 12 Maths Important Questions Chapter-by-Chapter Important Questions here:

CBSE Class 12 Maths Important Questions

Sr No Chapter No Chapter Name
1 Chapter 1 Relations and Functions
2 Chapter 2 Inverse Trigonometric Functions
3 Chapter 3 Matrices
4 Chapter 4 Determinants
5 Chapter 5 Continuity and Differentiability
6 Chapter 6 Application of Derivatives
7 Chapter 7 Integrals
8 Chapter 8 Application of Integrals
9 Chapter 9 Differential Equations
10 Chapter 10 Vector Algebra
11 Chapter 11 Three Dimensional Geometry
12 Chapter 12 Linear Programming
13 Chapter 13 Probability

Inverse Trigonometry Questions for Class 12 Maths Chapter 2

1. Show that tan ̄¹1+ + 1+1+1+ = 𝝅4 + 12cos ̄¹

Ans: To solve this question, substitute x²= cos , therefore,

tan ̄¹1+x² + 1+x²1+x²- 1+x² = tan ̄¹1+cos x  + 1-cos x1+cos x  – 1-cos x                   ..(1)

Using the trigonometric identities 1+cos x = 2 cosx2 and 1-cos x = 2 sinx2 on equation (1) we get,

tan ̄¹1+x² + 1+x²1+x²- 1+x²= tan ̄¹2 cos2+2 sin22 cos2-2 sin2       …..(2)

Taking 2 cos2 common from numerator and denominator of (2) we get,

tan ̄¹1+x² + 1+x²1+x²- 1+x²= tan ̄¹1+ tan 21- tan 2                            …..(3)

Now using the identity, tan ̄¹x+y1-xy=tan⁻¹(x)+tan⁻¹y on (3) we get,

tan ̄¹1+x² + 1+x²1+x²- 1+x²=tan⁻¹(x)+tan⁻¹tan2

⇒ tan ̄¹1+x² + 1+x²1+x²- 1+x²=𝝅4+12

Let this be known as equation (4).

Re-substituting x²= cos in (4)we get,

tan ̄¹1+x² + 1+x²1+x²- 1+x²=𝝅4+12cos ̄¹x²

Hence Proved.

1. Prove that 12sin ̄¹2x1+x²+12cos ̄¹1-y²1+y²=x+y1-xy

Ans. To solve this problem use the substitution,

x = tanθ

y = tanα

In the LHS of the given expression.

tan12sin⁻¹2×1+x²+12cos⁻¹1-y²1+y²= tan12sin⁻¹2tan1+tan²+12cos⁻¹2tan1+tan²

Let this be known as equation (1).

Using the trigonometric identity

2tan1+tan²=sin2

1-tan²1+tan²= cos2

From (1) we get,

tan 12sin⁻¹2×1+x²+12cos⁻¹1-y²1+y²= tan 12sin⁻¹sin2+12cos⁻¹cos 2

tan 12sin⁻¹2×1+x²+12cos⁻¹1-y²1+y²= tan +

Let this be known as equation (2).

Now using the trigonometric identity, tan x+y=tan x+tan y1-tan x tan y on equation (2), we get:

tan 12sin⁻¹2×1+x²+12cos⁻¹1-y²1+y²=tan +tan 1-tan tan                     ……(3)

Re-substituting =tan ̄¹x and =tan ̄¹y in equation(3) we get,

tan 12sin⁻¹2×1+x²+12cos⁻¹1-y²1+y² = tan(tan ̄¹x) + tan(tan ̄¹y)1-tan(tan ̄¹x)tan(tan ̄¹y)

tan 12sin ̄¹2×1+x²+12cos⁻¹1-y²1+y²=x+y1-xy

1. Show that tan ̄¹1+cos x + 1-cos x1+cos   – 1-cos x=𝝅4+x2,x𝝐[0,𝝅]

Ans. Using trigonometric identities

1+cos x = 2 cos x2 and 1-cos x = 2 sinx2 we get,

tan ̄¹1+cos x  + 1-cos x1+cos x  – 1-cos x = tan ̄¹2 cosx2+2 sin x2 1+cos x  – 1-cos x                ….(1)

Dividing the numerator and denominator of (1) by 2 cos x2 we get,

tan ̄¹1+cos x  + 1-cos x1+cos x  – 1-cos x = tan ̄¹ 1+ tanx21-tanx2                         .…(2)

Now using the identity, tan ̄¹x + y1-xy = tan ̄¹x+tan ̄¹(y) on the equation(2) we get

tan ̄¹1+cos x  + 1-cos x1+cos x  – 1-cos x = tan ̄¹(1)+tan ̄¹tanx2

𝝅4+x2

1. Show that

tan ̄¹xa²-x²=sin ̄¹xa=cos ̄¹a²-x²a

Ans. Using the trigonometric substitution x=a sin in tan ̄¹xa²-x²we get,

tan ̄¹xa²-x²=tan ̄¹a sin a²-a²sin²                                            ……..(1)

Taking out “a” common from the denominator of (1) and cancelling out with numerator we get

tan ̄¹xa²-x²=tan ̄¹sin 1- sin²                                             …….(2)

Using the trigonometric identity 1-sin²x = cos x in the denominator of (2) we get,

tan ̄¹xa²-x²=tan ̄¹sin cos

tan ̄¹xa²-x²=tan ̄¹tan()

Let it be known as equation (3).

Using the result tan ̄¹(tan x)=x in (3) we get

tan ̄¹xa²-x²=tan ̄¹                                                                              …….(4)

Let us now re-substitute

x=a sin

= sin ̄¹xa

In equation (4) we get,

tan ̄¹xa²-x²=sin ̄¹xa                                                           ……..(5)

Now, from x=a sin we get

sin =xa

Hence, using the trigonometric identity,

sin²+ cos²=1,  we get

xa²+cos²=1

cos²=1-

cos =a²-x²a²

Let this be known as equation (6).

Taking cos ̄¹ on both sides of (6) we get,

cos ̄¹cos =cos ̄¹a²-x²a²

=cos ̄¹a²-x²a²

From equation (4) we get,

tan ̄¹xa²-x²=cos ̄¹a²-x²a².

1. Solve the following to find x : cot ̄¹2x+ cot ̄¹3x=𝝅4

Ans: To solve this question, use the identity cot ̄¹xy – 1x + y=cot ̄¹(x)+cot ̄¹(y)

cot ̄¹2x+ cot ̄¹3x=cot ̄¹6x²-15x

Hence, cot ̄¹6x²-15x= 𝝅4                                                                             ……..(1)

Taking cot on both sides we get,

cot cot ̄¹6x²-15x=cot𝝅4

6x²-15x=cot𝝅4

Let it be known as equation (2).

Solving the equation (2) by substituting the principal value of cot𝝅4we get,

6x²-1=5x

6x²-5x-1=0

Simplifying it further we get,

6x²+(-6x+x)-1=0

6x+1x-1=0

x=1, –16

But, x= –16  is not possible. Therefore,

x=1.

1. Prove that

tan ̄¹13+tan ̄¹15+tan ̄¹17+tan ̄¹18= 𝝅4

Ans.  To solve this problem use the trigonometric identity,

tan ̄¹x+y1-xy=tan ̄¹x+tan ̄¹y in the LHS of the given expression

Hence,

tan ̄¹13+tan ̄¹15+tan ̄¹17+tan ̄¹18=tan ̄¹13+151- 115+tan ̄¹17+181- 156

tan ̄¹13+tan ̄¹15+tan ̄¹17+tan ̄¹18=tan ̄¹47+tan ̄¹311

Let this be known as equation (1).

Again, using the trigonometric identity tan ̄¹x+y1-xy=tan ̄¹x+tan ̄¹y on (1) we get,

tan ̄¹13+tan ̄¹15+tan ̄¹17+tan ̄¹18=tan ̄¹47+3111 – 1277

tan ̄¹13+tan ̄¹15+tan ̄¹17+tan ̄¹18=tan ̄¹6565

tan ̄¹13+tan ̄¹15+tan ̄¹17+tan ̄¹18=tan ̄¹1

Let this be known as equation (2).

But we know that,

tan𝝅4=1

tan ̄¹1= 𝝅4

Hence from equation (2) it is proved that

tan ̄¹13+tan ̄¹15+tan ̄¹17+tan ̄¹18= 𝝅4.

1. Evaluate: tan12cos ̄¹311

Ans.  To solve this question, use the substitution 12cos ̄¹311=x  in the LHS of the given expression. Therefore,

12cos ̄¹311=x

cos 2x=311

Let this be known as equation (1).

But we know that 1-tan²x1+tan²x= cos 2x. Hence from (1) we get,

1-tan²x1+tan²x=311                                                                       ……..(2)

Applying the rule of component and divided on equation (2) we get,

1-tan²x+1+tan²x1-tan²x-1-tan²x=3 + 113 – 11

2-2tan²x=3 + 113 – 11

Let this be known as equation (3).

Taking the reciprocal of equation (3) we get,

-tan²x=3 – 113 + 11

tan²x= 11 – 33 + 11

tan x=11 – 33+11

Let this be known as equation (4).

Now re-substituting  12cos ̄¹311=x  in (4) we get,

tan 12cos ̄¹311=11 – 33 + 11

Class 12 Maths Chapter 2 Important Questions – Inverse Trigonometric Functions

Introduction

The sine, cosine, tangent, secant, cosecant, and cotangent carry out opposite operations which are given by the inverse trigonometric functions. When two of the three side lengths are analyzed, they are used in a right triangle to evaluate the measurement of an angle.

Principal Inverse Trigonometric Functions With Domain and Range

 Inverse Trigonometric Function Domain Range sin -1, 1 -𝜋/2, 𝜋/2 cos -1, 1 0, 𝜋 tan R (-𝜋/2, 𝜋/2) cosec R -(-1,1) -𝜋/2, 𝜋/2-0 sec R (0, 𝜋 – 𝜋/2) cot R (0, 𝜋)

Properties of Inverse Trigonometric Functions

The properties of the inverse trigonometric functions are important for better understanding of the concepts in this chapter as well as to solve the mathematical equations. The properties of inverse trigonometric relations signify the relationship between all the inverse trigonometric functions like sine, cosine, secant, cosecant, cotangent and tangent.

1. sin-1 (1/x) = cosec-1 x, x1 or x1
2. cos-1 (1/x) = sec-1 x, x1 or x1
3. tan-1 (1/x) = cot-1 x, x > 0
4. sin-1 (–x) = – sin-1 x, x ∈
5. –1,1
6. –1,1
7. tan-1(–x) = – tan-1 x, x ∈ R
8. cosec-1 (–x) = – cosec-1 x, | x | 1
9. cos-1 (–x) = – cos-1 x, x ∈
10. –1,1
11. –1,1
12. sec-1 (–x) = – sec-1 x, | x | 1
13. cot-1 (–x) = – cot-1 x, x ∈ R
14.  sin-1 x + cos-1 x = 2 , x ∈
15. –1,1
16. –1,1
17.  tan-1 x + cot-1 x = 2 , x ∈ R
18.  cosec-1 x + sec-1 x = 2 , | x | ≥ 1
19.  tan-1 x + tan-1 y = tan-1 (x + y / 1 – xy) , xy < 1
20.  tan-1 x – tan-1 y = tan-1 (x – y / 1+ xy) , xy > – 1
21.  2tan-11 x = sin-1 (2x / 1 + x2) , | x | ≤ 1
22.  2tan-1 x = cos-1 (1 – x2 / 1 + x2) , x ≥ 0
23.  2tan-1 x = tan-1 (2x / 1 –  x2) , – 1 < x < 1

important questions of chapter 2 maths class 12 are made with a step-by-step approach by Extramarks subject matter experts.

Conclusion

Students can go through Class 12 Maths Chapter 2 Important Questions for their final exam preparations. Extramarks has prepared these questions with solutions for the Class 12 Maths Chapter 2 “Inverse Trigonometric Functions,” which includes all the necessary concepts as per the CBSE syllabus.

Q1.

opt.

ans.

$\begin{array}{l}\mathrm{We}\mathrm{have}\\ \mathrm{cot}\left[{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{3}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{5}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{7}}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{8}}\right]\\ =\mathrm{cot}\left[{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{5}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{7}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{3}+{\mathrm{tan}}^{–1}\frac{\mathrm{1}}{\mathrm{8}}\right]\\ =\mathrm{„}\mathrm{cot}\mathrm{„}\left[{\mathrm{tan}}^{1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1\frac{1}{5}Ã—\frac{1}{7}}\right)+{\mathrm{tan}}^{1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1\frac{1}{3}Ã—\frac{1}{8}}\right)\right]\\ =\mathrm{„}\mathrm{cot}\mathrm{„}\left[{\mathrm{tan}}^{1}\left(\frac{\frac{7+5}{35}}{1\frac{1}{35}}\right)+{\mathrm{tan}}^{1}\left(\frac{\frac{8+3}{24}}{1\frac{1}{24}}\right)\right]\\ =\mathrm{„}\mathrm{cot}\mathrm{„}\left[{\mathrm{tan}}^{1}\left(\frac{\frac{12}{35}}{\frac{34}{35}}\right)+{\mathrm{tan}}^{1}\left(\frac{\frac{11}{24}}{\frac{23}{24}}\right)\right]\\ =\mathrm{„}\mathrm{cot}\mathrm{„}\left[{\mathrm{tan}}^{1}\left(\frac{12}{34}\right)+{\mathrm{tan}}^{1}\left(\frac{11}{23}\right)\right]\\ =\mathrm{„}\mathrm{cot}\mathrm{„}\left[{\mathrm{tan}}^{1}\left(\frac{6}{17}\right)+{\mathrm{tan}}^{1}\left(\frac{11}{23}\right)\right]\\ =\mathrm{„}\mathrm{cot}\mathrm{„}+\left[{\mathrm{tan}}^{1}\left(\frac{\frac{138+187}{17Ã—23}}{\frac{39166}{17Ã—23}}\right)\right]\\ =\mathrm{„}\mathrm{cot}\mathrm{„}+\left[{\mathrm{tan}}^{1}\left(\frac{325}{325}\right)\right]\\ \\ =\mathrm{„}\mathrm{cot}\mathrm{„}+\left[{\mathrm{tan}}^{1}\left(1\right)\right]\\ =\mathrm{„}\mathrm{cot}\mathrm{„}\frac{\mathrm{Ï}}{4}\\ =\mathrm{„}1\end{array}$

Q2-

opt-

ans-

$\begin{array}{l}\mathrm{We}\mathrm{have}\\ {\mathrm{sin}}^{1}\left(1\mathrm{x}\right)2{\mathrm{sin}}^{1}\mathrm{x}\mathrm{„}=\frac{\mathrm{Ï}}{2}\\ \mathrm{„}{\mathrm{sin}}^{1}\mathrm{„}\left(1\mathrm{x}\right)\mathrm{„}=\mathrm{„}\frac{\mathrm{Ï}}{2}\mathrm{„}+\mathrm{„}2\mathrm{„}{\mathrm{sin}}^{1}\mathrm{x}\\ \mathrm{„}\left(1\mathrm{x}\right)\mathrm{„}=\mathrm{„}\mathrm{sin}\mathrm{„}\left(\frac{\mathrm{Ï}}{2}+2{\mathrm{sin}}^{1}\mathrm{x}\right)\\ \mathrm{„}\left(1\mathrm{x}\right)\mathrm{„}=\mathrm{„}\mathrm{cos}\mathrm{„}\left(2\mathrm{„}{\mathrm{sin}}^{1}\mathrm{x}\right)\\ \mathrm{„}{\mathrm{cos}}^{1}\left(1\mathrm{x}\right)\mathrm{„}=\mathrm{„}2{\mathrm{sin}}^{1}\mathrm{x}\\ \mathrm{„}{\mathrm{sin}}^{1\mathrm{„}}\sqrt{1{\left(1\mathrm{x}\right)}^{2}}=\mathrm{„}{\mathrm{sin}}^{1}\left(2\mathrm{x}\sqrt{1{\mathrm{x}}^{2}}\right)\\ \sqrt{1{\left(1\mathrm{x}\right)}^{2}}\mathrm{„}=\mathrm{„}\left(2\mathrm{x}\mathrm{„}\sqrt{1{\mathrm{x}}^{2}}\right)\\ „„„\mathrm{On}\mathrm{squaring}\mathrm{both}\mathrm{sides}\\ 1–{\left(1\mathrm{x}\right)}^{2}=4{\mathrm{x}}^{2}\left(1{\mathrm{x}}^{2}\right)\\ \mathrm{„}11{\mathrm{x}}^{2}+2\mathrm{x}=\mathrm{„}4{\mathrm{x}}^{2}4{\mathrm{x}}^{4}\\ \mathrm{„}4{\mathrm{x}}^{4}5{\mathrm{x}}^{2}+2\mathrm{x}\mathrm{„}=\mathrm{„}0\\ \mathrm{„}\mathrm{x}\mathrm{„}\left(4{\mathrm{x}}^{3}5\mathrm{x}+2\right)\mathrm{„}=\mathrm{„}0\\ ’\mathrm{„}\mathrm{x}=\mathrm{}0\mathrm{}\mathrm{or}\\ 4{\mathrm{x}}^{3}5\mathrm{x}+2=\mathrm{„}0\\ \mathrm{Put}\mathrm{x}=...-1 0 1 2.\dots \mathrm{we}\mathrm{get}\mathrm{no}\mathrm{real}\mathrm{vlue}\mathrm{of}\mathrm{x}\\ \mathrm{satisfy}\mathrm{above}\mathrm{cubic}\mathrm{equation}\mathrm{in}\left[1\mathrm{„}1\right]\\ \mathrm{Hence}\mathrm{x}= 0\mathrm{is}\mathrm{only}\mathrm{solution}.\end{array}$

Q3-

opt-

ans-

$\begin{array}{l}{\mathrm{cos}}^{’1}\left(\mathrm{cos}\frac{13\mathrm{}}{6}\right)\mathrm{„}=\mathrm{„}\frac{13\mathrm{}}{6}\\ \mathrm{But}\frac{13\mathrm{}}{6}\mathrm{„}ˆ‰\mathrm{„}\left[0\mathrm{„}\mathrm{}\right]\\ \mathrm{cos}„\left(\frac{13\mathrm{}}{6}\right)\mathrm{„}=\mathrm{cos}\left(2\mathrm{}+\frac{\mathrm{}}{6}\right)\mathrm{„}=\mathrm{„}\mathrm{cos}\mathrm{„}\left(\frac{\mathrm{}}{6}\right)\\ ˆ´„„{\mathrm{cos}}^{1}\mathrm{„}\left(\mathrm{cos}\frac{13\mathrm{}}{6}\right)\mathrm{„}=\mathrm{„}\frac{\mathrm{}}{6}\end{array}$

Q.4

Option –

$\begin{array}{l}\left(\mathrm{i}\right){\mathrm{sin}}^{–1}\left(\frac{–1}{\mathrm{2}}\right)\mathrm{}=\mathrm{}\mathrm{y}\\ \mathrm{sin}\mathrm{}\mathrm{y}\mathrm{}=\mathrm{}\frac{1}{2}\\ \mathrm{sin}\mathrm{}\mathrm{y}\mathrm{}=\mathrm{}\mathrm{sin}\mathrm{}\frac{\mathrm{}}{}\end{array}$

Ans.

Q5-

Opt-

Ans-

$\begin{array}{l}{\mathrm{tan}}^{1}\left(\mathrm{1}\right)+{\mathrm{cos}}^{–1}\left(\frac{–1}{\mathrm{2}}\right)+{\mathrm{sin}}^{–1}\left(\frac{–1}{\mathrm{2}}\right)\\ {\mathrm{tan}}^{1}\left(\mathrm{tan}\frac{\mathrm{}}{4}\right)+{\mathrm{cos}}^{–1}\left(\mathrm{cos}\frac{\mathrm{}\mathrm{}}{3}\right)+{\mathrm{sin}}^{–1}\left(\mathrm{sin}\frac{\mathrm{}}{6}\right)\\ =\mathrm{}\frac{\mathrm{}}{4}+{\mathrm{cos}}^{–1}\left(\mathrm{cos}\left(\mathrm{}\frac{\mathrm{}\mathrm{}}{3}\right)\right)+{\mathrm{sin}}^{–1}\left[\mathrm{sin}\left(\frac{\mathrm{}}{6}\right)\right]\\ \mathrm{we}\mathrm{know}\mathrm{range}\mathrm{of}\\ {\mathrm{tan}}^{1}\mathrm{}†’\mathrm{}\left(\frac{’\mathrm{}}{2}\mathrm{}\frac{\mathrm{}\mathrm{}}{2}\right)\\ {\mathrm{cos}}^{1}\mathrm{}†’\mathrm{}\left[0\mathrm{}\mathrm{}\right]\mathrm{}\mathrm{and}\mathrm{}{\mathrm{sin}}^{1}\mathrm{}†’\mathrm{}\left[\frac{\mathrm{}}{2}\mathrm{}\frac{\mathrm{}\mathrm{}}{2}\right]\\ \mathrm{}{\mathrm{tan}}^{ˆ1}\left(1\right)+{\mathrm{cos}}^{1}\left(\frac{’1}{2}\right)\mathrm{}+\mathrm{}{\mathrm{sin}}^{1}\left(\frac{1}{2}\right)\\ =\frac{\mathrm{}}{4}\mathrm{}+\mathrm{}\frac{2\mathrm{}}{3}\frac{\mathrm{}}{6}\\ \mathrm{}=\mathrm{}\frac{3\mathrm{}+8\mathrm{}’2\mathrm{}}{12}\mathrm{}=\mathrm{}\frac{9\mathrm{}}{12}\mathrm{}=\mathrm{}\frac{3\mathrm{}}{4}\end{array}$

CBSE Class 12 Maths Important Questions

1. Prove that sin-1 (3/5) – sin-1(8/17) = cos-1 (84/85).

Let sin-1 (3/5) = a and sin-1 (8/17) = b

Thus, we can write sin a = 3/5 and sin b = 8/17

Now, find the value of cos a and cos b

To find cos a:

Cos a = √[1 – sin2a]

= √[1 – (3/5)2 ]

= √[1 – (9/25)]

= √[(25-9)/25]

= 4/5

Thus, the value of cos a = 4/5

To find cos b:

Cos b= √[1 – sin2b]

= √[1 – (8/17)2 ]

= √[1 – (64/289)]

= √[(289-64)/289]

= 15/17

Thus, the value of cos b = 15/17

We know that cos (a- b) = cos a cos b + sin a sin b

Now, substitute the values for cos a, cos b, sin a and sin b in the formula, we get:

cos (a – b) = (4/5)x (15/17) + (3/5)x(8/17)

cos (a – b) = (60 + 24)/(17x 5)

cos (a – b) = 84/85

(a – b) = cos-1 (84/85)

Substituting the values of a and b sin-1 (3/5)- sin-1 (8/7) = cos-1 (84/85)

Hence proved.

2. Find the value of cot (tan-1 α + cot-1 α).

Given that: cot (tan-1 𝛂 + cot-1 𝛂)

= cot (𝝅/𝟐) (since, tan-1 x + cot-1 x = 𝜋/2)

= cot (180°/2) (we know that cot 90° = 0)

= cot (90°)

= 0

Therefore, the value of cot (tan-1 α + cot-1 α) is 0.