Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Exercise 2.2 | Set 1

Exercise 2.2 | Set 1 focuses on the properties and applications of inverse trigonometric functions. This set of problems is designed to deepen students' understanding of these functions, their domains, ranges, and various identities. Inverse trigonometric functions, also known as arc functions, are crucial in many areas of mathematics and physics, particularly in solving equations involving trigonometric functions. This exercise challenges students to manipulate these functions, solve equations, and prove identities, thereby enhancing their analytical and problem-solving skills. The concepts covered here form a foundation for more advanced topics in calculus and complex analysis.

Prove the following

Question 1. 3sin^-1x = sin^-1(3x - 4x³), x∈[-1/2, 1/2] 

Solution:

Let us take x = sinθ, so θ = sin^-1x

Substitute the value of x in the equation present on R.H.S. 

The equation becomes sin^-1(3sinθ - 3sin³θ) 

We know, sin3θ = 3sinθ - 4sin³θ

So , sin^-1(3sinθ - 3sin³θ) = sin^-1(sin3θ) 

By the property of inverse trigonometry we know, sin(sin^-1(θ)) = θ   

So, sin^-1(sin3θ) = 3θ 

And we know θ = sin^-1x   

So, 3θ = 3sin^-1x = L.H.S

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

Solution:

Let us take x = cosθ, so θ = cos^-1x

Substitute value of x in the equation present on R.H.S. 

The equation becomes cos^-1(4cos³θ - 3cosθ) 

We know, cos3θ = 4cos³θ - 3cosθ  

So, cos^-1(4cos³θ - 3cosθ) = cos^-1(cos3θ) 

By the property of inverse trigonometry we know, cos(cos^-1(θ)) = θ    

So, cos^-1(cos3θ) = 3θ 

And we know θ = cos^-1x     

So, 3θ = 3cos^-1x = L.H.S

Question 3. tan^{-1}\frac{2}{11} + tan^{-1}\frac{7}{24} = tan^{-1}\frac{1}{2}  

Solution:

We know,  tan^{-1}x + tan^{-1}y = tan^{-1}\frac{x+y}{1-xy}  

Now put x = 2/11 and y = 7/24

So,  tan^{-1}\frac{2}{11} + tan^{-1}\frac{7}{24} = tan^{-1}\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}.\frac{7}{24}}  

=tan^{-1}\frac{\frac{48}{264}+\frac{77}{264}}{1-\frac{14}{264}}  

=tan^{-1}\frac{\frac{48+77}{264}}{\frac{264-14}{264}}  

=tan^{-1}\frac{\frac{48+77}{264}}{\frac{264-14}{264}}  

=tan^{-1}\frac{125}{250}  

=tan^{-1}\frac{1}{2}        

= R.H.S

Question 4.  2tan^{-1}\frac{1}{2} + tan^{-1}\frac{1}{7} = tan^{-1}\frac{31}{17}  

Solution:

We have to first write 2tan^-1x in terms of  tan^-1x

We know that 2tan^-1x = tan^{-1}\frac{2x}{1-x^2}    

Put x = 1/2 in the above formula

So,  2tan^{-1}\frac{1}{2} = tan^{-1}\frac{2.\frac{1}{2}}{1-(\frac{1}{2})^2}

= tan^{-1}\frac{1}{1-(\frac{1}{4})^2}

= tan^{-1}\frac{1}{\frac{4-1}{4}}

= tan^{-1}\frac{1}{\frac{3}{4}}

= tan^{-1}\frac{4}{3}

Now we can replace 2tan^{-1}\frac{1}{2}   with  tan^{-1}\frac{4}{3}

So equation in L.H.S become  tan^{-1}\frac{4}{3} + tan^{-1}\frac{1}{7}

We know ,  tan^{-1}x + tan^{-1}y = tan^{-1}\frac{x+y}{1-xy}           

So,   tan^{-1}\frac{4}{3} + tan^{-1}\frac{1}{7} = tan^{-1}\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}.\frac{1}{7}}  

=tan^{-1}\frac{\frac{28}{21}+\frac{3}{21}}{1-\frac{4}{21}}  

=tan^{-1}\frac{\frac{28+3}{21}}{\frac{21-4}{21}}  

=tan^{-1}\frac{\frac{31}{21}}{\frac{17}{21}}  

=tan^{-1}\frac{31}{17}          

= R.H.S

Write the following functions in simplest forms: 

Question 5.  tan^{-1}\frac{\sqrt{1+x^2}-1}{x} , x ≠ 0 

Solution:

Let us assume that x = tanθ, so θ = tan^-1x 

Substitute the value of x in question. 

So equation becomes tan^{-1}\frac{\sqrt{1+tan^2θ}-1}{tanθ}

We know that, 1 + tan²θ = sec²θ 

Replacing 1 + tan²θ with sec²θ in the equation tan^{-1}\frac{\sqrt{1+tan^2θ}-1}{tanθ}

So equation becomes, tan^{-1}\frac{\sqrt{sec^2θ}-1}{tanθ}

= tan^{-1}\frac{secθ-1}{tanθ}

We know, tanθ = sinθ/cosθ and sec = 1/cosθ 

Replacing value of tanθ and secθ in  tan^{-1}\frac{secθ-1}{tanθ}

= tan^{-1}\frac{\frac{1}{cosθ}-1}{\frac{sinθ}{cosθ}}

= tan^{-1}\frac{\frac{1-cosθ}{cosθ}}{\frac{sinθ}{cosθ}}

= tan^{-1}\frac{1-cosθ}{sinθ}

We know, 1 - cosθ = 2sin²θ/2​ and sinθ = 2sinθ/2cosθ/2  

So the equations after replacing above value becomes tan^{-1}\frac{2sin^2\frac{θ}{2}}{2sin\frac{θ}{2}cos\frac{θ}{2}}

= tan^{-1}\frac{sin\frac{θ}{2}}{cos\frac{θ}{2}}

We know \frac{\frac{sinθ}{2}}{\frac{cosθ}{2}} = tan\frac{θ}{2}        

= tan^{-1}tan\frac{θ}{2}

= θ/2           [tan^-1(tanθ) = θ]

= 1/2 tan^-1x            [θ = tan^-1x]   

Question 6. tan^{-1}\frac{1}{\sqrt{x^2-1}}  , |x| > 1

Solution:

Let us assume that x = cosecθ, so θ = cosec ^-1x 

Substitute the value of x in question with cosecθ 

=tan^{-1}\frac{1}{\sqrt{cosec^2θ-1}}

We know that, 1 + cot²θ = cosec²θ, so cosec²θ = 1 - cot²θ  

=tan^{-1}\frac{1}{\sqrt{cot^2θ}}

=tan^{-1}\frac{1}{cotθ}

= tan^-1(tanθ)          [1/cotθ = tanθ]  

= θ           [tan^-1(tanθ) = θ] 

= cosec⁻¹x          [θ = cosec⁻¹x]

= π/2 ​- sec⁻¹x         [cosec⁻¹x + sec⁻¹x = π/2​]

Question 7. tan^{-1}(\sqrt{\frac{1-cosx}{1+cosx}}), 0<x<\pi

Solution:

We know, 1 - cosx = 2sin²x/2 and 1 + cosx = 2cos²x/2   

 Substituting above formula in question

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

=tan^{-1}(\frac{sin\frac{x}{2}}{cos\frac{x}{2}})

= tan^-1(tanx/2)         [\frac{sin\theta}{cos\theta} = tan\theta]

 = x/2         [tan^-1(tanθ) = θ] 

Question 8. tan^{-1}(\frac{cosx-sinx}{cosx+sinx}), \frac{-\pi}{4}<x<\frac{3\pi}{4}

Solution:

Divide numerator and denominator by cosx

= tan^{-1}(\frac{\frac{cosx-sinx}{cosx}}{\frac{cosx+sinx}{cosx}})

= tan^{-1}(\frac{\frac{cosx}{cosx}-\frac{sinx}{cosx}}{\frac{cosx}{cosx}+\frac{sinx}{cosx}})

= tan^{-1}(\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}})

We know, \frac{sinx}{cosx} = tanx

= tan^{-1}(\frac{1-tanx}{1+tanx})

This can also be written as tan^{-1}(\frac{1-tanx}{1+1.tanx})         - (1)

We know  tan^{-1}x-tan^{-1}y = tan^{-1}\frac{x-y}{1+x.y}            - (2)

On comparing equation (1) and (2) we can say that x = 1 and y = tan^-1x

So we can say that tan^{-1}(\frac{1-tanx}{1+1.tanx}) = tan^{-1}1 - tan^{-1}x

= π/4​ - tan⁻¹x          [tan⁻¹1 = π/4​]

Question 9. tan^{-1}\frac{x}{\sqrt{a^2-x^2}} , |x|<a

Solution:

Let us assume that x = asinθ, so θ = sin ^-1x/a

Substitute the value of x in question.

= tan^{-1}\frac{asin\theta}{\sqrt{a^2-(asin\theta)^2}}

= tan^{-1}\frac{asin\theta}{\sqrt{a^2-a^2sin^2\theta}}

Taking a² common from denominator

= tan^{-1}\frac{asin\theta}{\sqrt{a^2(1-sin^2\theta)}}

We know that, sin²θ + cos²θ = 1, so 1 - sin²θ = cos²θ 

= tan^{-1}\frac{asin\theta}{\sqrt{a^2cos^2\theta}}

= tan^{-1}\frac{asin\theta}{acos\theta}

= tan^{-1}\frac{sin\theta}{cos\theta}

= tan^-1(tanθ)          [sinθ/cosθ = tanθ]

 = θ        

= sin^-1x/a 

Question 10. tan^{-1}(\frac{3a^2x-x^3}{a^3-3ax^2}),a>0;\frac{-a}{\sqrt{3}}<x<\frac{a}{\sqrt{3}}

Solution:

Let us assume that x = atanθ, so θ = tan ^-1x/a

Substitute the value of x in question

 = tan^{-1}(\frac{3a^2(atan\theta)-(atan\theta)^3}{a^3-3a(atan\theta)^2})

= tan^{-1}(\frac{3a^3tan\theta-a^3tan^3\theta}{a^3-3a^3tan^2\theta})

Taking a³common from numerator and denominator

= tan^{-1}(\frac{a^3(3tan\theta-tan^3\theta)}{a^3(1-3tan^2\theta)})

= tan^{-1}(\frac{3tan\theta-tan^3\theta}{1-3tan^2\theta})

We know tan3\theta = \frac{3tan\theta-tan^3\theta}{1-3tan^2\theta}

So, tan^{-1}(\frac{3tan\theta-tan^3\theta}{1-3tan^2\theta})= tan^{-1}(tan3\theta)

= 3θ           [ tan^-1(tanθ) = θ]

= 3tan ^-1x/a

Summary

Exercise 2.2 | Set 1 provides a comprehensive exploration of inverse trigonometric functions, challenging students to apply their knowledge in various contexts. Through solving equations, proving identities, and evaluating complex expressions, students develop a deeper intuition for these functions and their properties. The problems in this set often require a combination of algebraic manipulation, trigonometric identities, and logical reasoning. By working through these questions, students not only reinforce their understanding of inverse trigonometric functions but also enhance their overall mathematical problem-solving skills. This exercise serves as a crucial bridge between basic trigonometry and more advanced mathematical concepts, preparing students for future studies in calculus, complex analysis, and physics where these functions play a significant role.