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

Last Updated : 03 Apr, 2024

### Question 11. tanâˆ’1[2cos(2sinâˆ’11/2â€‹)]

Solution:

Let us assume that sinâˆ’11/2 = x

So, sinx = 1/2

Therefore, x = Ï€â€‹/6 = sinâˆ’11/2

Therefore, tanâˆ’1[2cos(2sinâˆ’11/2â€‹)] =  tanâˆ’1[2cos(2 * Ï€â€‹/6)]

= tanâˆ’1[2cos(Ï€â€‹/3)]

Also, cos(Ï€/3â€‹) = 1/2â€‹

Therefore, tanâˆ’1[2cos(Ï€â€‹/3)] = tanâˆ’1[(2 * 1/2)]

= tanâˆ’1[1] = Ï€â€‹/4

### Question 12. cot(tanâˆ’1a + cotâˆ’1a)

Solution:

We know, tanâˆ’1x + cotâˆ’1x = Ï€â€‹/2

Therefore, cot(tanâˆ’1a + cotâˆ’1a) = cot(Ï€â€‹/2) =0

### Question 13.  [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)â€‹[2(tanâˆ’1x + tanâˆ’1y)]

= tan[tanâˆ’1x + tanâˆ’1y]

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

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

= (x + y)/(1 – xy)

### Question 14. If sin(sinâˆ’11/5â€‹ + cosâˆ’1x) = 1 then find the value of x

Solution:

sinâˆ’11/5â€‹ + cosâˆ’1x = sinâˆ’11

We know, sinâˆ’11 = Ï€/2

Therefore, sinâˆ’11/5â€‹ + cosâˆ’1x = Ï€/2

sinâˆ’11/5â€‹ = Ï€/2 – cosâˆ’1x

Since, sinâˆ’1xâ€‹ + cosâˆ’1x = Ï€/2

Therefore, Ï€/2 – cosâˆ’1x = sinâˆ’1x

sinâˆ’11/5â€‹ = sinâˆ’1x

So, x = 1/5

### Question 15. 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, tanâˆ’1x + tanâˆ’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/âˆš2, -1/âˆš2

### Question 16. sin âˆ’ 1(sin2Ï€/3â€‹)

Solution:

We know that sinâˆ’1(sinÎ¸) = Î¸ when Î¸ âˆˆ [-Ï€/2, Ï€/2], but [Tex]\frac{2 \pi}{3} > \frac{\pi}{2}[/Tex]

So, sin âˆ’ 1(sin2Ï€/3â€‹) can be written as [Tex]sin^{-1}[sin(\pi-\frac{2\pi}{3})][/Tex]

sin âˆ’ 1(sinÏ€/3â€‹)  here [Tex]\frac{-\pi}{2}<\frac{\pi}{3}<\frac{\pi}{2}[/Tex]

Therefore, sin âˆ’ 1(sinÏ€/3â€‹) = Ï€/3

### Question 17. tanâˆ’1(tan3Ï€/4â€‹)

Solution:

We know that tanâˆ’1(tanÎ¸) = Î¸ when [Tex]\theta \epsilon(\frac{-\pi}{2},\frac{\pi}{2}) [/Tex] but [Tex]\frac{3 \pi}{4} > \frac{\pi}{2}[/Tex]

So, tanâˆ’1(tan3Ï€/4â€‹) can be written as tanâˆ’1(-tan(-3Ï€/4)â€‹)

= tanâˆ’1[-tan(Ï€ – Ï€/4â€‹)]

= tanâˆ’1[-tan(Ï€/4â€‹)]

= –tanâˆ’1[tan(Ï€/4â€‹)]

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

### Question 18. [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

### (i) 7Ï€/6    (ii) 5Ï€/6    (iii)Ï€/3    (iv)Ï€/6

Solution:

We know that cosâˆ’1(cosÎ¸) = Î¸, Î¸ âˆˆ [0, Ï€]

cosâˆ’1(cosÎ¸) = Î¸, Î¸ âˆˆ [0, Ï€]

Here, 7Ï€/6 > Ï€

So, cosâˆ’1(cos7Ï€/6â€‹) can be written as cosâˆ’1(cos(-7Ï€/6)â€‹)

= cosâˆ’1[cos(2Ï€ – 7Ï€/6â€‹)]      [cos(2Ï€ + Î¸) = Î¸]

= cosâˆ’1[cos(5Ï€/6â€‹)]       where 5Ï€/6 âˆˆ  [0, Ï€]

Therefore, cosâˆ’1[cos(5Ï€/6â€‹)] = 5Ï€/6

### (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 = -Ï€/6â€‹

Therefore, sin[Ï€/3â€‹ – (-Ï€/6â€‹)]

= sin[Ï€/3â€‹ + (Ï€/6â€‹)]

= sin[3Ï€/6]

= sin[Ï€/2]

= 1

### (i) Ï€    (ii) -Ï€/2    (iii)0    (iv)2âˆš3

Solution:

We know, cot(âˆ’x) = âˆ’cotx

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

= tan-13 + cot-13

Since, tan-1x + cot-1x = Ï€/2

Tan-13 + cot-13 = -Ï€/2

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