NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions

NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions will give you an idea of the technics used in solving the different types of questions based on the inverse trigonometric functions. The NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions are solved and scrutinized by a team of experts. The purpose of these NCERT solutions is to help students in clearing their doubts so that they could achieve excellent marks in the exam.

NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions

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Q1.3Sin^-1x = Sin^-1(3x – 4x³) ; x ∈ [1/2,-1/2]

Ans. Considering Sin^-1x = θ

sin θ = x

Taking RHS ,we have

Sin^-1(3x – 4x³)

Putting the value of x = sin θ

Sin^-1(3sin θ – 4sin ³θ)[since sin 3θ =3sin θ – 4sin ³θ]

Sin^-1 (sin 3θ)

= 3θ = 3Sin^-1x = LHS

Hence proved

Q2.3cos^-1x = cos^-1(3x – 4x³) ; x ∈ [1/2,1]

Ans. Considering cos^-1x = θ

cos θ = x

Taking RHS ,we have

cos^-1(3x – 4x³)

Putting the value of x = cos θ

cos^-1(3cos³ θ – 4cos θ)[since cos 3θ = 4cos³θ -3cos θ]

cos^-1 (cos 3θ)

= 3θ = 3cos^-1x = LHS

Hence proved

Q3. Prove that

tan^-12/11 + tan^-17/24= tan^-11/2

Taking the LHS

tan^-12/11 + tan^-17/24

Applying the formula

$tan^{-1}x +tan^{-1}y=tan^{-1}\left ( \frac{x+y}{1-xy} \right )$

$=tan^{-1}\left ( \frac{2/11+7/24}{1-2/11 \times 7/24} \right )$

$=tan^{-1}\left [ \frac{(48+77)/264}{(264-14)/264} \right ]$

= Tan^-1(125/250)

= Tan^-11/2 = RHS, Hence proved

NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions

Q4. Prove that

2tan^-1(1/2) +tan^-1(1/7) =tan^-1(31/17)

Taking the LHS

2tan^-1(1/2) +tan^-1(1/7)

Reawritting it as following

tan^-1(1/2)+tan^-1(1/2) +tan^-1(1/7)

Applying the formula

$=tan^{-1}1/2+tan^{-1}\left [ \frac{1/2+1/7}{1-1/14} \right ]$

$=tan^{-1}1/2+tan^{-1}\left ( \frac{9/14}{13/14} \right )$

= tan^-1(1/2) +tan^-1(9/13)

Again applying the same formula

$=tan^{-1}\left ( \frac{1/2+9/13}{1-9/26} \right )$

$=tan^{-1}\left ( \frac{31/26}{17/26} \right )$

= tan^-1(31/17) = RHS, Hence proved

Q5.Write the function in the simplest form

$tan^{-1}\frac{\sqrt{1+x^{2}}-1}{x}$

Ans. Considering the value of x = tan θ

We get

$=tan ^{-1}\frac{\sqrt{1+tan^{2}\Theta }-1}{tan\Theta }$

$=tan^{-1}\frac{\sqrt{sec^{2}\Theta } -1}{tan\Theta }$

$=tan^{-1}\frac{(sec\Theta -1)}{tan\Theta }$

$=tan^{-1}\frac{(1-cos\Theta )}{sin\Theta }$

$=tan^{-1}\frac{2sin^{2}\Theta /2}{2sin\Theta/2.cos\Theta /2 }$

$=tan^{-1}\frac{sin\Theta /2}{cos\Theta /2}$

$=tan^{-1}(tan\Theta/2 )$

= θ/2

Q6. Write the function in the simplest form

$tan^{-1}\frac{1}{\sqrt{x^{2}-1}}, \left | x \right |> 1$

Ans. Considering x = cosec θ

$=\frac{1}{\sqrt{cosec^{2}\Theta-1 }}$

$=tan^{-1}\frac{1}{\sqrt{cot^{2}\Theta } }$

$=tan^{-1}\frac{1}{cot\Theta }$

Tan^-1(tan θ) = θ

Since x = cosec θ⇒θ = cosec^-1x

Q7. Write the function in the simplest form

$tan^{-1}\sqrt{\frac{1-cosx}{1+cosx}},x<\pi$

Ans. The given function is

$tan^{-1}\sqrt{\frac{1-cosx}{1+cosx}}$

Since 1 – cos x = 2sin²(x/2) and 1 + cos x = 2cos²(x/2)

$=tan^{-1}\sqrt{\frac{2sin^{2}\left ( x/2 \right )}{2cos^{2}\left ( x/2 \right )}}$

$=tan^{-1}\left ( \frac{sin\: x/2}{cos\: x/2} \right )$

=tan^-1(tan x/2) = x/2

NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions

Q8. Write the function in the simplest form

$tan^{-1}\left [ \frac{cosx-sinx}{cosx+sinx} \right ],0<x<\pi$

Ans. The given function is

$tan^{-1}\left [ \frac{cosx-sinx}{cosx+sinx} \right ]$

For writting the given function in the form of tan,dividing the numerator and denominator by cos x

$=tan^{-1}\left [ \frac{cosx/cosx-sinx/cosx}{cosx/cosx+sinx/cosx} \right ]$

$=tan^{-1}\left [\frac{1-tanx}{1+tanx} \right ]$

Writing it in the following form

$tan\left ( A-B \right )=\frac{tanA-tanB}{1+tanA.tanB}$

$=tan^{-1}[\frac{tan\pi /4-tanx}{tan\pi /4+tan\pi /4.tanx}],\because tan\pi /4=1$

= tan^-1[tan(π/4 – x)]

= π/4 – x

Q9. Write the function in the simplest form

$tan^{-1}\frac{x}{\sqrt{x^{2}-a^{2}}},\left | x \right |<a$

Ans. The given function is

$tan^{-1}\frac{x}{\sqrt{x^{2}-a^{2}}},$

Considering x = a sin θ, we have

$tan^{-1}\frac{asin\Theta }{\sqrt{a^{2}.sin^{2}\Theta -a^{2}}},$

$=tan^{-1}\left [ \frac{asin\Theta }{a\sqrt{sin^{2 }\Theta -1}} \right ]$

$=tan^{-1}\left [ \frac{sin\Theta }{\sqrt{sin^{2 }\Theta -1}} \right ]$

$=tan^{-1}\left (\frac{sin\Theta }{cos\Theta } \right )$

= tan^-1(tan θ) = θ

Since, x = a sin θ, hence sin θ = x/a

θ = sin^-1(x/a)

Q10. Write the function in the simplest form

$tan^{-1}\left [ \frac{3a^{2}x-x^{3}}{a^{3}-3ax^{2}} \right ],a>0, \left ( -\frac{a}{\sqrt{3}}\leq x\leq \frac{a}{\sqrt{3}} \right )$

Ans.The given function is

$tan^{-1}\left [ \frac{3a^{2}x-x^{3}}{a^{3}-3ax^{2}} \right ]$

Considering x = a tan θ, we have

$=tan^{-1}\left [ \frac{3a^{2}.atan\Theta -a^{3}tan^{3}\Theta }{a^{3}-3a.a^{2}.tan^{2}\Theta } \right ]$

$=tan^{-1}\left [ \frac{a^{3}\left ( 3tan\Theta -tan^{3}\Theta \right )}{a^{3}\left ( 1-3tan^{2}\Theta \right )} \right ]$

$=tan^{-1}\left [ \frac{3tan\Theta -tan^{3}\Theta }{1-3tan^{2}\Theta } \right ]$

= tan^-1(tan 3θ) = 3θ

Since x = a tan θ ⇒ tan θ = x/a , θ = tan^-1(x/a)

Therefore 3θ = 3tan^-1(x/a)

Find the value of each of the following:

$Q11.\: tan^{-1}\left [ 2cos\left ( 2sin^{-1}\frac{1}{2} \right ) \right ]$

Ans.

$tan^{-1}\left [ 2cos\left ( 2sin^{-1}\frac{1}{2} \right ) \right ]$

$=tan^{-1}\left [ 2cos\left ( 2sin^{-1}.sin\frac{\pi }{6} \right ) \right ]$

$=tan^{-1}\left [ 2cos\left ( 2.\frac{\pi }{6} \right ) \right ]$

$=tan^{-1}\left [ 2cos\frac{\pi }{3} \right ]$

$=tan^{-1}\left [ 2.\frac{1}{2} \right ]$

= tan^-11 ⇒ π/4

cot π/2 =0

NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions

Q12. cot ( sin^-1α + cos^-1 α)

Ans. We are given

cot ( sin^-1α + cos^-1 α)

Since, sin^-1x + cos^-1 x) = π/2

cot π/2 = 0

$Q13.tan\frac{1}{2}\left [ sin^{-1}\frac{2x}{1+x^{2}}+cos^{-1}\frac{1-y^{2}}{1+y^{2}} \right ],\left | x \right |<1,y>0,and \: xy<1$

Ans. We are given the function

$tan\frac{1}{2}\left [ sin^{-1}\frac{2x}{1+x^{2}}+cos^{-1}\frac{1-y^{2}}{1+y^{2}} \right ]$

Considering x = tan A and y = tan B

$=tan\frac{1}{2}\left [ sin^{-1}\frac{2tanA}{1+tan^{2}A}+cos^{-1}\frac{1-tan^{2}B}{1+tan^{2}B} \right ]$

$=tan \frac{1}{2}\left [ sin^{-1}\frac{2tanA}{sec^{2}A} +cos^{-1}\frac{1-tan^{2}B}{sec^{2}B}\right ]$

$=tan \frac{1}{2}\left [ sin^{-1}\frac{2sinA/cosA}{1/cos^{2}A} +cos^{-1}\frac{1-sin^{2}B/cos^{2}B}{1/cos^{2}B}\right ]$

$=tan\frac{1}{2}\left [ sin^{-1} .2sinA.cosA+cos^{-1}\left ( cos^{2}B-sin^{2} B\right )\right ]$

$=tan\frac{1}{2}\left [ sin^{-1} .sin 2A+cos^{-1}cos2B\right ]$

$=tan\frac{1}{2}\left [ 2A+2B\right ]$

= tan (A + B)

$=\frac{tanA+tanB}{1-tanA.tanB}$

Since x = tan A and y = tan B

$=\frac{x+y}{1-xy}$

Q14.If sin[ sin^-1(1/5) + cos^-1x ] =1, then find the value of x

Ans. We are given that

sin[ sin^-1(1/5) + cos^-1x ] =1

Substituting 1=Sin π/2

sin[ sin^-1(1/5) + cos^-1x ] =sin π/2

sin^-1(1/5) + cos^-1x = π/2

Since sin^-1A + cos^-1A= π/2

Therefore

cos^-1x = π/2 – sin^-1(1/5) =cos^-11/5

x = 1/5

$Q15.\: If \: tan^{-1}\frac{x-1}{x-2}+tan^{-1}\frac{x+1}{x+2}=\frac{\pi }{4},then\: find\: the\: value\: of\: x$

Ans. We are given that

$\: tan^{-1}\frac{x-1}{x-2}+tan^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$

Applying the following formula

$tan^{-1}x +tan^{-1}y=tan^{-1}\frac{x+y}{1-xy}$

$tan^{-1}\frac{\left ( x-1 \right )/\left ( x-2 \right ) +\left ( x+1 \right )/\left ( x+2 \right ) }{1-\left ( x-1 \right )/\left ( x-2 \right ) .\left ( x+1 \right )/x+2 }=\frac{\pi }{4}$

$tan^{-1}\frac{\left ( x-1 \right )\left ( x+2 \right )+\left ( x+1 \right )\left ( x-2 \right )}{\left ( x-2 \right )\left ( x+2 \right )-\left ( x-1 \right )\left ( x+1 \right )}=\frac{\pi }{4}$

$tan^{-1}\frac{x^{2}+x-2+x^{2}-x-2}{x^{2}-4-x^{2}+1}=\frac{\pi }{4}$

$tan^{-1}\frac{2x^{2}-4}{-3}=\frac{\pi }{4}$

$\frac{2x^{2}-4}{-3}=tan\frac{\pi }{4}$

2x² -4 = -3

2x² = 1⇒ x = ±1/√2

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

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Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

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Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

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NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals