## NCERT Solutions Class 12th Maths Chapter 2: Inverse Trigonometric Functions## Exercise 2.1
1. sin
Let sin sin y = -1/2 sin y = -sin π/6 sin y = sin (-π/6) We know that the range of the principal value branch of sin Hence, the principal value of sin 2. cos
Let cos cos y = √3/2 co y = cos π/6 We know that the range of the principal value branch of cos Hence, the principal value of cos 3. cosec
Let cosec cosec y = 2 cosec y = cosec π/6 We know that the range of the principal value branch of cosec Hence, the principal value of cosec 4. tan
Let tan tan y = -√3 tan y = -tan π/3 tan y = tan (-π/3) We know that the range of the principal value branch of tan Hence, the principal value of tan 5. cos
Let cos cos y = -1/2 cos y = -cos π/3 cos y = cos (π - π/3) cos y = cos (2π/3) We know that the range of the principal value branch of cos Hence, the principal value of cos 6. tan
Let tan tan y = -1 tan y = -tan π/4 tan y = tan (-π/4) We know that the range of the principal value branch of tan Hence, the principal value of tan 7. sec
Let sec sec y = 2/√3 sec y = sec π/6 We know that the range of the principal value branch of sec Hence, the principal value of sec 8. cot
Let cot cot y = √3 cot y = cot π/6 We know that the range of the principal value branch of cot Hence, the principal value of cot 9. cos
Let cos cos y = -1/√2 cos y = -cos π/4 cos y = cos (π - π/4) cos y = cos (3π/4) We know that the range of the principal value branch of cos Hence, the principal value of cos 10. cosec
Let cosec cosec y = -√2 cosec y = -cosec π/4 cosec y = cosec (-π/4) We know that the range of the principal value branch of cosec Hence, the principal value of cosec
11. tan
Let tan tan x = 1 tan x = tan π/4 So, tan Let cos cos y = -1/2 cos y = -cos π/3 cos y = cos (π - π/3) cos y = cos 2π/3 So, cos Let sin sin z = -1/2 sin z = -sin π/6 sin z = sin (-π/6) sin Therefore, tan = (3π + 8π - 2π)/12 = 9π/12 = 3π/4 12. cos
Let cos cos x = 1/2 cos x = cos π/3 So, cos Let sin sin y = 1/2 sin y = sin π/6 So, sin Therefore, cos = π/3 + π/3 = 2π/3 13. If sin **0 ≤ y ≤ π****-π/2 ≤ y ≤ π/2****0 < y < π****-π/2 < y < π/2**
sin sin y = x We know that the range of the principal value branch of sin Therefore, -π/2 ≤ y ≤ π/2. 14. Find the value of tan **π****-π/3****π/3****2π/3**
Let tan tan x = √3 tan x = tan π/3 We know that the range of the principal value branch of tan Therefore, tan Let sec sec y = -2 sec y = -sec π/6 sec y = sec (π - π/6) sec y = sec 2π/3 We know that the range of the principal value branch of sec Therefore, sec Hence, tan ## Exercise 2.2
1. 3 sin
Let x = sin θ sin Take RHS = sin = sin = sin = 3θ = 3 sin = LHS Hence, proved. 2. 3 cos
Let x = cos θ cos Take RHS = cos = cos = cos = 3θ = 3 cos = LHS Hence, proved.
3. tan
Let x = tan θ tan tan = tan Using 1 - cos 2A = 2 sin tan = tan = tan = θ/2 = (tan 4. tan
tan Using 1 - cos 2A = 2 sin tan = tan = tan = x/2 5. tan
tan Divide both the numerator and denominator by cos x in [(cos x - sin x)/(cos x + sin x)] tan = tan = tan = tan Using tan = tan = tan = π/4 - x 6. tan
Let x = a sin θ sin θ = x/a sin tan = tan = tan = tan = tan = θ = sin 7. tan
Let x = a tan θ tan θ = x/a tan tan = tan = tan = tan = tan = tan = 3θ = 3 tan
8. tan
Let sin sin x = 1/2 sin x = sin π/6 Therefore, sin tan = tan = tan = tan = π/4 9. tan [sin
Let x = tan θ tan Now, sin Using sin 2A = 2 tan A/(1 + tan sin = sin = 2θ = 2 tan Let y = tan ϕ tan Now, cos Using cos 2A = (1 - tan cos = cos = 2ϕ = 2 tan So, tan [sin = tan [2 tan = tan [2(tan = tan (tan Using tan tan (tan = tan [tan = (x + y)/(1 - xy)
10. sin
We know that sin 2π/3 ∉ [-π/2, π/2] So, sin = sin Now, π/3 ∈ [-π/2, π/2]. Therefore, sin 11. tan
We know that tan 3π/4 ∉ (-π/2, π/2) So, tan = tan = tan = tan Now, -π/4 ∈ (-π/2, π/2). Therefore, tan = -π/4 12. tan (sin
Let sin sin x = 3/5 sin cos cos x = √(1 - sin cos x = √(1 - 9/25) cos x = √(16/25) cos x = 4/5 tan x = sin x/cos x tan x = (3/5)/(4/5) tan x = 3/4 x = tan So, sin tan Therefore, tan (sin = tan [tan Using tan tan [tan = tan [tan = tan [tan = tan [tan = tan [tan = 17/6 13. cos **7π/6****5π/6****π/3****π/6**
We know that cos 7π/6 ∉ [0, π] So, cos = cos = cos Now, 5π/6 ∈ [0, π]. Therefore, cos = cos = 5π/6 Hence, the correct answer is (B). 14. sin (π/3 - sin **1/2****1/3****1/4****1**
Let sin sin x = -1/2 sin x = -sin π/6 sin x = sin (-π/6) sin We know that the range of the principal value of branch of sin sin (π/3 - sin = sin (π/3 + π/6) = sin (3π/6) = sin π/2 = 1 Hence, the correct answer is (D). 15. tan **π****-π/2****0****2√3**
We know that the range of the principal value branch of tan tan = tan = tan = π/3 - cot = π/3 - 5π/6 = -3π/6 = -π/2 Hence, the correct answer is (B). ## Miscellaneous Exercise
1. cos
We know that cos 13π/6 ∉ [0, π] So, cos = cos = cos Now, π/6 ∈ [0, π]. Therefore, cos = π/6 2. tan
We know that tan 7π/6 ∉ (-π/2, π/2) So, tan = tan = tan = tan = tan Now, π/6 ∈ (-π/2, π/2). Therefore, tan = π/6
3. 2 sin
Let sin sin x = 3/5 cos cos cos x = √(1 - (3/5) cos x = √(1 - 9/25) cos x = √(16/25) cos x = 4/5 tan x = sin x/cos x tan x = (3/5)/(4/5) tan x = 3/4 x = tan So, Now, LHS = 2 sin = 2 tan Using 2 tan 2 tan = tan = tan = tan Hence, proved. 4. sin
Let sin sin x = 8/17 cos cos cos x = √(1 - sin cos x = √(1 - (8/17) cos x = √(1 - 64/289) cos x = √(22/289) cos x = 15/17 tan x = sin x/cos x tan x = (8/17)/(15/17) tan x = 8/15 x = tan So, Let sin sin y = 3/5 cos cos cos y = √(1 - (3/5) cos y = √(1 - 9/25) cos y = √(16/25) cos y = 4/5 tan y = sin y/cos y tan y = (3/5)/(4/5) tan y = 3/4 y = tan So, Now, LHS = sin = tan Using tan tan = tan = tan = tan = tan Hence, proved. 5. cos
Let cos cos x = 4/5 sin sin sin x = √(1 - (4/5) sin x = √(1 - 16/25) sin x = √(9/25) sin x = 3/5 tan x = sin x/cos x tan x = (3/5)/(4/5) tan x = 3/4 x = tan So, Let cos cos y = 12/13 sin sin sin y = √(1 - (12/13) sin y = √(1 - 144/169) sin y = √(25/169) sin y = 5/13 tan y = sin y/cos y tan y = (5/13)/(12/13) tan y = 5/12 y = tan So, Let cos cos z = 33/65 sin sin sin z = √(1 - (33/65) sin z = √(1 - 1089/4225) sin z = √(3136/4225) sin z = 56/65 tan z = sin z/cos z tan z = (56/65)/(33/65) tan z = 56/33 z = tan So, Now, LHS = cos = tan Using tan tan = tan = tan = tan = tan = cos Hence, proved. 6. cos
Let cos cos x = 12/13 sin sin sin x = √(1 - (12/13) sin x = √(1 - 144/169) sin x = √(25/169) sin x = 5/13 tan x = sin x/cos x tan x = (5/13)/(12/13) tan x = 5/12 x = tan So, Let sin sin y = 3/5 cos cos cos y = √(1 - (3/5) cos y = √(1 - 9/25) cos y = √(16/25) cos y = 4/5 tan y = sin y/cos y tan y = (3/5)/(4/5) tan y = 3/4 y = tan So, Let sin sin z = 56/65 cos cos cos z = √(1 - sin cos z = √(1 - (56/65) cos z = √(1 - 3136/4225) cos z = √(1089/4225) cos z = 33/65 tan z = sin z/cos z tan z = (56/65)/(33/65) tan z = 56/33 z = tan So, Now, LHS = cos = tan Using tan tan = tan = tan = tan = tan = sin Hence, proved. 7. tan
Let sin sin x = 5/13 x = sin cos cos cos x = √(1 - sin cos x = √(1 - (5/13) cos x = √(1 - 25/169) cos x = √(144/169) cos x = 12/13 tan x = sin x/cos x tan x = (5/13)/(12/13) tan x = 5/12 x = tan So, Let cos cos y = 3/5 sin sin sin y = √(1 - cos sin y = √(1 - (3/5) sin y = √(1 - 9/25) sin y = 4/5 tan y = sin y/cos y tan y = (4/5)/(3/5) tan y = 4/3 y = tan So, 4/3Now, RHS = sin = tan Using tan tan = tan = tan = tan = tan Hence, proved.
8. tan
Let x = tan √x = tan θ θ = tan Therefore, RHS = 1/2 cos = 1/2 cos Using (1 - tan = 1/2 cos = 1/2 × 2θ = θ = tan Hence, proved. 9. cot
Rationalise [√(1 + sin x) + √(1 - sin x)]/[√(1 + sin x) - √(1 - sin x)] [√(1 + sin x) + √(1 - sin x)]/[√(1 + sin x) - √(1 - sin x)] × [√(1 + sin x) - √(1 - sin x)]/[√(1 + sin x) - √(1 - sin x)] = [√(1 + sin x) + √(1 - sin x)] = [(1 + sin x) + (1 - sin x) + 2√(1 + sin x)(1 - sin x)]/[1 + sin x - 1 + sin x] = [2 + 2√(1 - sin = 2[1 + √(1 - sin = [1 + √cos = [1 + cos x]/sin x = [2 cos = (cos x/2)/(sin x/2) = cot x/2 Therefore, LHS = cot = cot = x/2 = RHS Hence, proved. 10. tan
Let x = cos 2θ 2θ = cos θ = 1/2 cos Now, LHS = tan = tan = tan = tan = tan = tan = tan Using tan tan = tan = π/4 - θ = π/4 - 1/2 cos Hence, proved.
11. 2 tan
2 tan Using 2 tan tan 2 cos x/(1 - cos 2 cos x/sin cos x/sin cos x/sin cos x/sin x = 1 cot x = 1 x = cot
12. tan
tan Using tan tan π/4 = 1/2 tan π/4 = (tan π/4 = 3/2 tan tan x = tan π/6
13. sin (tan **x/√(1 - x**^{2})**1/√(1 - x**^{2})**1/√(1 + x**^{2})**x/√(1 + x**^{2})
Let tan tan y = x Using tan tan Taking sin on both sides, sin (tan sin (tan Hence, (D) is the correct answer. 14. sin **0, 1/2****1, 1/2****0****1/2**
Given: sin Let x = sin y y = sin sin sin sin Taking sin on both sides, sin [sin 1 - sin y = sin (π/2 + 2y) 1 - sin y = cos 2y Using cos 2A = 1 - 2 sin 1 - sin y = 1 - 2 sin 2 sin 2x x (2x - 1) = 0
2x - 1 = 0 ⇒ But x ≠ 1/2 as it won't satisfy the given equation. Therefore, x = 0 is the solution of the given equation. Hence, (C) is the correct answer. |