TOPICS

Exercise - 2.1

Inverse Trigonometry

**Question-1 :-** Find the principal value of sin^{-1}(-1/2).

Let sin^{-1}(1/2) = θ. sin θ = 1/2 sin θ = sin(π/6) {Range = [-π/2, π/2]} θ = π/6 The principal value of sin^{-1}(1/√2) = π/6

**Question-2 :-** Find the principal value of cos^{-1}(√3/2).

Let cos^{-1}(√3/2) = θ. cos θ = √3/2 cos θ = cot(π/6) {Range = [0, π]} θ = π/6 The principal value of cos^{-1}(√3/2) = π/6

**Question-3 :-** Find the principal value of cosec^{-1}(2).

Let cosec^{-1}(2) = θ. cosec θ = 2 cosec θ = coesec(π/6) {Range = [0, π]} θ = π/6 The principal value of cosec^{-1}(2) = π/6

**Question-4 :-** Find the principal value of tan^{-1}(-√3).

Let tan^{-1}(-√3) = θ. tan θ = -√3 tan θ = -tan(π/3) {Range = [-π/2, π/2]} tan θ = tan(-π/3) θ = -π/3 The principal value of tan^{-1}(-√3) = -π/3

**Question-5 :-** Find the principal value of cos^{-1}(-1/2).

Let cos^{-1}(-1/2) = θ. cos θ = -1/2 cos θ = -cos(π/3) {Range = [0, π]} cos θ = cos(π - π/3) cos θ = cos(2π/3) θ = 2π/3 The principal value of cos^{-1}(-1/2) = 2π/3

**Question-6 :-** Find the principal value of tan^{-1}(-1).

Let tan^{-1}(-1) = θ. tan θ = -1 tan θ = -tan(π/4) {Range = [-π/2, π/2]} tan θ = tan(-π/4) θ = -π/4 The principal value of tan^{-1}(-√3) = -π/4

**Question-7 :-** Find the principal value of sec^{-1}(2/√3).

Let sec^{-1}(2/√3) = θ. sec θ = 2/√3 sec θ = sec(π/6) {Range = [0, π]} θ = π/6 The principal value of sec^{-1}(2/√3) = π/6

**Question-8 :-** Find the principal value of cot^{-1}(√3).

Let cot^{-1}(√3) = θ. cot θ = √3 cot θ = cot(π/6) {Range = [0, π]} θ = π/6 The principal value of cot^{-1}(√3) = π/6

**Question-9 :-** Find the principal value of cos^{-1}(-1/√2).

Let cos^{-1}(-1/√2) = θ. cos θ = -1/√2 cos θ = -cos(π/4) {Range = [0, π]} cos θ = cos(π - π/4) cos θ = cos(3π/4) θ = 3π/4 The principal value of cos^{-1}(-1/√2) = 3π/4

**Question-10 :-** Find the principal value of cosec^{-1}(-√2).

Let cosec^{-1}(-√2) = θ. cosec θ = -√2 cosec θ = -cosec(π/4) {Range = [-π/2, π/2]} cosec θ = cosec(-π/4) θ = -π/4 The principal value of cosec^{-1}(-√2) = -π/4

**Question-11 :-** Find the value of tan^{-1}(1) + cos^{-1}(-1/2) + sin^{-1}(-1/2).

tan^{-1}(1) + cos^{-1}(-1/2) + sin^{-1}(-1/2) Let tan^{-1}(1) = θ₁. tan θ₁ = 1 tan θ₁ = tan(π/4) {Range = [-π/2, π/2]} θ₁ = π/4 Let cos^{-1}(-1/2) = θ₂. cos θ₂ = -1/2 cos θ₂ = -cos(π/3) {Range = [0, π]} cos θ₂ = cos(π - π/3) cos θ₂ = cos(2π/3) θ₂ = 2π/3 Let sin^{-1}(-1/2) = θ₃. sin θ₃ = -1/2 sin θ₃ = -sin(π/6) {Range = [-π/2, π/2]} sin θ₃ = sin(-π/6) θ₃ = π/6 Now, θ₁ + θ₂ + θ₃ = π/4 + 2π/3 + π/6 = (3π + 8π + 2π)/12 = 13π/12

**Question-12 :-** Find the value of cos^{-1}(1/2) + 2sin^{-1}(1/2).

cos^{-1}(1/2) + 2sin^{-1}(1/2) Let cos^{-1}(1/2) = θ₁. cos θ₁ = 1/2 cos θ₁ = cos(π/3) {Range = [0, π]} θ₁ = π/3 Let sin^{-1}(1/2) = θ₂ sin θ₂ = 1/2 sin θ₂ = sin(π/6) {Range = [-π/2, π/2]} θ₂ = π/6 Now, θ₁ + 2θ₂ = π/3 + 2 x π/6 = π/3 + π/3 = 2π/3

**Question-13 :-** If sin^{-1} x = y, then

(A) 0 ≤ y ≤ π (B) -π/2 ≤ y ≤ π/2 (C) 0 < y < π (D) -π/2 < y < π/2

(B) -π/2 ≤ y ≤ π/2, because sin^{-1}x range is [-π/2, π/2].

**Question-14 :-** tan^{-1}(√3) - sec^{-1}(-2) is equal to

(A) π (B) -π/3 (C) π/3 (D) 2π/3

tan^{-1}(√3) - sec^{-1}(-2) Let tan^{-1}(√3) = θ₁. tan θ₁ = √3 tan θ₁ = tan(π/3) {Range = [-π/2, π/2]} θ₁ = π/3 Let sec^{-1}(-2) = θ₂. sec θ₂ = -1/2 sec θ₂ = -sec(π/3) {Range = [0, π]} sec θ₂ = sec(π - π/3) sec θ₂ = sec(2π/3) θ₂ = 2π/3 Now, θ₁ - θ₂ = π/3 - 2π/3 = (π - 2π)/3 = -π/3 Option (B).

CLASSES