Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Exercise 2.2 | Set 2
Last Updated :
03 Apr, 2024
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​ = Ï€/2 – cos−1x
Since, sin−1x​ + cos−1x = π/2
Therefore, Ï€/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(Ï€ – Ï€/4​)]
= tan−1[-tan(π/4​)]
= –tan−1[tan(Ï€/4​)]
= – 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
Question 19. cos−1(cos7π/6​) is equal to
(i) 7Ï€/6 (ii) 5Ï€/6 (iii)Ï€/3 (iv)Ï€/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(2Ï€ – 7Ï€/6​)] [cos(2Ï€ + 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[Ï€/3​ – (-Ï€/6​)]
= sin[π/3​ + (π/6​)]
= sin[3Ï€/6]
= sin[Ï€/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 = -Ï€/2
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