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

Last Updated : 03 Apr, 2024
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Content of this article has been merged with Chapter 2 Inverse Trigonometric Functions – Exercise 2.2 as per the revised syllabus of NCERT.

Find the values of each of the following: 

Question 11. 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 12. cot(tan−1a + cot−1a) 

Solution:

We know, tan−1x + cot−1x = π​/2

Therefore, cot(tan−1a + cot−1a) = cot(π​/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)​[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 14. 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

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, 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.

Question 16. 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 17. 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 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

Question 19.  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 20. [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 21. [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



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