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Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Exercise 2.2 | Set 1
  • Difficulty Level : Expert
  • Last Updated : 05 Apr, 2021

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-1

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 -1

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

Chapter 2 Inverse Trigonometric Functions – Exercise 2.2 | Set 2

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