Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Exercise 2.2

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

Write the following functions in simplest forms: 

Question 3.  [Tex]tan^{-1}\frac{\sqrt{1+x^2}-1}{x} , x \neq 0 [/Tex]

Solution:

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

Substitute the value of x in question. 

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

We know that, 1 + tan²θ = sec²θ 

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

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

[Tex]= tan^{-1}\frac{secθ-1}{tanθ}[/Tex]

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

Replacing value of tanθ and secθ in [Tex] tan^{-1}\frac{secθ-1}{tanθ}[/Tex]

[Tex]= tan^{-1}\frac{\frac{1}{cosθ}-1}{\frac{sinθ}{cosθ}}[/Tex]

[Tex]= tan^{-1}\frac{\frac{1-cosθ}{cosθ}}{\frac{sinθ}{cosθ}}[/Tex]

[Tex]= tan^{-1}\frac{1-cosθ}{sinθ}[/Tex]

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

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

[Tex]= tan^{-1}\frac{sin\frac{θ}{2}}{cos\frac{θ}{2}}[/Tex]

We know [Tex]\frac{\frac{sinθ}{2}}{\frac{cosθ}{2}} = tan\frac{θ}{2}       [/Tex] 

[Tex]= tan^{-1}tan\frac{θ}{2}[/Tex]

= θ/2           [tan^-1(tanθ) = θ]

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

Question 4. [Tex]tan^{-1}(\sqrt{\frac{1-cosx}{1+cosx}}), 0<x<\pi[/Tex]

Solution:

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

 Substituting above formula in question

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

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

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

 = x/2         [tan^-1(tanθ) = θ] 

Question 5. [Tex]tan^{-1}(\frac{cosx-sinx}{cosx+sinx}), \frac{-\pi}{4}<x<\frac{3\pi}{4}[/Tex]

Solution:

Divide numerator and denominator by [Tex]cosx[/Tex]

[Tex]= tan^{-1}(\frac{\frac{cosx-sinx}{cosx}}{\frac{cosx+sinx}{cosx}})[/Tex]

[Tex]= tan^{-1}(\frac{\frac{cosx}{cosx}-\frac{sinx}{cosx}}{\frac{cosx}{cosx}+\frac{sinx}{cosx}})[/Tex]

[Tex]= tan^{-1}(\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}})[/Tex]

We know, [Tex]\frac{sinx}{cosx} = tanx[/Tex]

[Tex]= tan^{-1}(\frac{1-tanx}{1+tanx})[/Tex]

This can also be written as [Tex]tan^{-1}(\frac{1-tanx}{1+1.tanx})       [/Tex]  – (1)

We know  [Tex]tan^{-1}x-tan^{-1}y = tan^{-1}\frac{x-y}{1+x.y}      [/Tex]      – (2)

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

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

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

Question 6. [Tex]tan^{-1}\frac{x}{\sqrt{a^2-x^2}} , |x|<a[/Tex]

Solution:

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

Substitute the value of x in question.

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

[Tex]= tan^{-1}\frac{asin\theta}{\sqrt{a^2-a^2sin^2\theta}}[/Tex]

Taking a² common from denominator

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

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

[Tex]= tan^{-1}\frac{asin\theta}{\sqrt{a^2cos^2\theta}}[/Tex]

[Tex]= tan^{-1}\frac{asin\theta}{acos\theta}[/Tex]

[Tex]= tan^{-1}\frac{sin\theta}{cos\theta}[/Tex]

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

 = θ        

= sin^-1x/a 

Question 7. [Tex]tan^{-1}(\frac{3a^2x-x^3}{a^3-3ax^2}),a>0;\frac{-a}{\sqrt{3}}<x<\frac{a}{\sqrt{3}}[/Tex]

Solution:

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

Substitute the value of x in question

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

[Tex]= tan^{-1}(\frac{3a^3tan\theta-a^3tan^3\theta}{a^3-3a^3tan^2\theta})[/Tex]

Taking a³common from numerator and denominator

[Tex]= tan^{-1}(\frac{a^3(3tan\theta-tan^3\theta)}{a^3(1-3tan^2\theta)})[/Tex]

[Tex]= tan^{-1}(\frac{3tan\theta-tan^3\theta}{1-3tan^2\theta})[/Tex]

We know [Tex]tan3\theta = \frac{3tan\theta-tan^3\theta}{1-3tan^2\theta}[/Tex]

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

= 3θ           [ tan^-1(tanθ) = θ]

= 3tan ^-1x/a

Find the values of each of the following: 

Question 8. tan⁻¹[2cos(2sin⁻¹1/2​)]

Solution:

Let us assume that sin⁻¹1/2 = x

So, sinx = 1/2

Therefore, x = π​/6 = sin⁻¹1/2

Therefore, tan⁻¹[2cos(2sin⁻¹1/2​)] =  tan⁻¹[2cos(2 * π​/6)]

= tan⁻¹[2cos(π​/3)]

Also, cos(π/3​) = 1/2​

Therefore, tan⁻¹[2cos(π​/3)] = tan⁻¹[(2 * 1/2)]

= tan⁻¹[1] = π​/4 

Question 9.  [Tex]tan\frac{1}{2}[sin^{-1}\frac{2x}{1+x^2}+cos^{-1}\frac{1-y^2}{1+y^2}],|x|<1,y>0,xy<1[/Tex]

Solution:

We know, 2tan^-1x = [Tex]sin^{-1}\frac{2 x}{1+x^2}   [/Tex] and 2tan^-1y =  [Tex]cos^{-1}[\frac{1 – y^2 }{1+y^2}][/Tex]

[Tex]\therefore tan\frac{1}{2}[sin^{-1}\frac{2x}{1+x^2}+cos^{-1}\frac{1-y^2}{1+y^2}]   [/Tex] 

= tan(1/2)​[2(tan⁻¹x + tan⁻¹y)]

= tan[tan⁻¹x + tan⁻¹y]

Also, tan⁻¹x + tan⁻¹y = [Tex]tan^{-1}\frac{x+y}{1-xy}[/Tex]

Therefore, tan[tan⁻¹x + tan⁻¹y] = [Tex]tan[tan^{-1}\frac{x+y}{1-xy}][/Tex]

= (x + y)/(1 – xy)

Question 10. sin ^{− 1}(sin2π/3​)  

Solution:

We know that sin⁻¹(sinθ) = θ when θ ∈ [-π/2, π/2], but [Tex]\frac{2 \pi}{3} > \frac{\pi}{2}[/Tex]

So, sin ^{− 1}(sin2π/3​) can be written as [Tex]sin^{-1}[sin(\pi-\frac{2\pi}{3})][/Tex]

 sin ^{− 1}(sinπ/3​)  here [Tex]\frac{-\pi}{2}<\frac{\pi}{3}<\frac{\pi}{2}[/Tex]

Therefore, sin ^{− 1}(sinπ/3​) = π/3

Question 11. tan⁻¹(tan3π/4​)

Solution:

We know that tan⁻¹(tanθ) = θ when [Tex]\theta \epsilon(\frac{-\pi}{2},\frac{\pi}{2}) [/Tex] but [Tex]\frac{3 \pi}{4} > \frac{\pi}{2}[/Tex]

So, tan⁻¹(tan3π/4​) can be written as tan⁻¹(-tan(-3π/4)​)

= tan⁻¹[-tan(π – π/4​)]

= tan⁻¹[-tan(π/4​)]

= –tan⁻¹[tan(π/4​)]

= – π/4 where [Tex]\frac{-\pi}{4} \epsilon(\frac{-\pi}{2},\frac{\pi}{2})[/Tex]

Question 12. [Tex]tan(sin^{-1}\frac{3}{5} + cot^{-1}\frac{3}{2})[/Tex]

Solution:

Let us assume [Tex]sin^{-1}\frac{3}{5}  [/Tex] = x , so sinx = 3/5 

We know, [Tex]cosx = \sqrt{1-sin^2x}[/Tex]

[Tex]\therefore cosx = \sqrt{1-(\frac{3}{5})^2}[/Tex]

[Tex]cosx = \sqrt{1-\frac{9}{25}}[/Tex]

[Tex]cosx = \sqrt{\frac{25-9}{25}}[/Tex]

[Tex]cosx = \sqrt{\frac{16}{25}}[/Tex]

cosx = 4/5

We know, [Tex]tanx = \frac{sinx}{cosx}[/Tex]

So, [Tex]tanx = \frac{\frac{3}{5}}{\frac{4}{5}}[/Tex]

tanx = 3/4

Also, [Tex]tan^{-1}\frac{1}{x} = cot^{-1}x[/Tex]

Hence, [Tex]tan(sin^{-1}\frac{3}{5} + cot^{-1}\frac{3}{2}) = tan(tan^{-1}\frac{3}{4}+tan^{-1}\frac{2}{3})[/Tex]

tan^-1x + tan^-1y = [Tex]tan^{-1}\frac{x+y}{1-xy}[/Tex]

So, [Tex]tan(tan^{-1}\frac{3}{4}+tan^{-1}\frac{2}{3}) = tan(tan^{-1}\frac {\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}.\frac{2}{3}})[/Tex]

[Tex]= tan(tan^{-1}\frac{\frac{9+8}{12}}{\frac{12-6}{12}})[/Tex]

[Tex]= tan(tan^{-1}\frac{17}{6})[/Tex]

= 17/6

Question 13.  cos⁻¹(cos7π/6​) is equal to

(i) 7π/6    (ii) 5π/6    (iii)π/3    (iv)π/6

Solution:

 We know that cos⁻¹(cosθ) = θ, θ ∈ [0, π]

cos⁻¹(cosθ) = θ, θ ∈ [0, π]

Here, 7π/6 > π 

So, cos⁻¹(cos7π/6​) can be written as cos⁻¹(cos(-7π/6)​)

= cos⁻¹[cos(2π – 7π/6​)]      [cos(2π + θ) = θ]

= cos⁻¹[cos(5π/6​)]       where 5π/6 ∈  [0, π]

  Therefore, cos⁻¹[cos(5π/6​)] = 5π/6

Question 14. [Tex]sin[\frac{\pi}{3} – sin^{-1}(-\frac{1}{2} )][/Tex]

(i) 1/2    (ii) 1/3   (iii) 1/4    (iv) 1

Solution:

Let us assume sin^-1(-1/2)= x, so sinx = -1/2 

Therefore, x = -π/6​

Therefore, sin[π/3​ – (-π/6​)]

= sin[π/3​ + (π/6​)]

= sin[3π/6]

= sin[π/2]

= 1

Question 15. [Tex]tan^{-1}\sqrt{3} – cot^{-1}(-\sqrt{3})  [/Tex] is equal to

(i) π    (ii) -π/2    (iii)0    (iv)2√3

Solution:

We know, cot(−x) = −cotx

Therefore, tan^-13 – cot^-1(-3) = tan^-13 – [-cot^-1(3)]

= tan^-13 + cot^-13

Since, tan^-1x + cot^-1x = π/2

Tan^-13 + cot^-13 = -π/2

Deleted Questions

[Tex] tan^{-1}\frac{2}{11} + tan^{-1}\frac{7}{24} = tan^{-1}\frac{1}{2}  [/Tex]

Solution:

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

Now put x = 2/11 and y = 7/24

So, [Tex] 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}}  [/Tex]

[Tex]=tan^{-1}\frac{\frac{48}{264}+\frac{77}{264}}{1-\frac{14}{264}}  [/Tex]

[Tex]=tan^{-1}\frac{\frac{48+77}{264}}{\frac{264-14}{264}}  [/Tex]

[Tex]=tan^{-1}\frac{\frac{48+77}{264}}{\frac{264-14}{264}}  [/Tex]

[Tex]=tan^{-1}\frac{125}{250}  [/Tex]

[Tex]=tan^{-1}\frac{1}{2}        [/Tex]

= R.H.S

 [Tex] 2tan^{-1}\frac{1}{2} + tan^{-1}\frac{1}{7} = tan^{-1}\frac{31}{17}  [/Tex]

Solution:

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

We know that 2tan^-1x = [Tex]tan^{-1}\frac{2x}{1-x^2}   [/Tex] 

Put x = 1/2 in the above formula

So,  [Tex]2tan^{-1}\frac{1}{2} = tan^{-1}\frac{2.\frac{1}{2}}{1-(\frac{1}{2})^2}[/Tex]

[Tex]= tan^{-1}\frac{1}{1-(\frac{1}{4})^2}[/Tex]

[Tex]= tan^{-1}\frac{1}{\frac{4-1}{4}}[/Tex]

[Tex]= tan^{-1}\frac{1}{\frac{3}{4}}[/Tex]

[Tex]= tan^{-1}\frac{4}{3}[/Tex]

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

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

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

So,  [Tex] 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}}  [/Tex]

[Tex]=tan^{-1}\frac{\frac{28}{21}+\frac{3}{21}}{1-\frac{4}{21}}  [/Tex]

[Tex]=tan^{-1}\frac{\frac{28+3}{21}}{\frac{21-4}{21}}  [/Tex]

[Tex]=tan^{-1}\frac{\frac{31}{21}}{\frac{17}{21}}  [/Tex]

[Tex]=tan^{-1}\frac{31}{17}         [/Tex] 

= R.H.S

 [Tex]tan^{-1}\frac{1}{\sqrt{x^2-1}} [/Tex] , |x| > 1

Solution:

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

Substitute the value of x in question with [Tex]cosecθ [/Tex]

[Tex]=tan^{-1}\frac{1}{\sqrt{cosec^2θ-1}}[/Tex]

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

[Tex]=tan^{-1}\frac{1}{\sqrt{cot^2θ}}[/Tex]

[Tex]=tan^{-1}\frac{1}{cotθ}[/Tex]

= 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​]

 cot(tan⁻¹a + cot⁻¹a) 

Solution:

We know, tan⁻¹x + cot⁻¹x = π​/2

Therefore, cot(tan⁻¹a + cot⁻¹a) = cot(π​/2) =0

If sin(sin⁻¹1/5​ + cos⁻¹x) = 1 then find the value of x

Solution:

sin⁻¹1/5​ + cos⁻¹x = sin⁻¹1

We know, sin⁻¹1 = π/2

Therefore, sin⁻¹1/5​ + cos⁻¹x = π/2

sin⁻¹1/5​ = π/2 – cos⁻¹x

Since, sin⁻¹x​ + cos⁻¹x = π/2

Therefore, π/2 – cos⁻¹x = sin⁻¹x

sin⁻¹1/5​ = sin⁻¹x

So, x = 1/5

If [Tex]tan^{-1}\frac{x-1}{x-2} + tan^{-1}\frac{x+1}{x+2} = \frac{\pi}{4}   [/Tex] , then find the value of x

Solution:

We know, tan⁻¹x + tan⁻¹y = [Tex]tan^{-1}\frac{x+y}{1-xy}[/Tex]

[Tex]tan^{-1}\frac{x-1}{x-2} + tan^{-1}\frac{x+1}{x+2} =tan^{-1}\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2}.\frac{x+1}{x+2}} = \frac{\pi}{4}[/Tex]

[Tex]tan^{-1}\frac{\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)}}{\frac{(x-2)(x+2)-(x+1)(x-1)}{(x-2)(x+2)}} = \frac{\pi}{4}[/Tex]

[Tex]tan^{-1}\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x+1)(x-1)} = \frac{\pi}{4}[/Tex]

[Tex]tan^{-1}(\frac{x^2+2x-x-2+x^2-2x+x-2}{x^2-4-x^2+1}) = \frac{\pi}{4}[/Tex]

[Tex]tan^{-1}(\frac{x^2+x-2+x^2-x-2}{-3}) = \frac{\pi}{4}[/Tex]

[Tex]tan^{-1}(\frac{2x^2-4}{-3}) = \frac{\pi}{4}[/Tex]

[Tex]\frac{2x^2-4}{-3} = tan(\frac{\pi}{4})[/Tex]

[Tex]\frac{2x^2-4}{-3} = 1[/Tex]

2x² – 4 = -3

2x² – 4 + 3 = 0

2x² – 1 = 0

x² = 1/2

x = 1/√2, -1/√2

Find the values of each of the expressions in Exercises 16 to 18.