Question-1 :- Find the value of cos-1 (cos 13π/6).
Solution :-cos-1 (cos 13π/6) Here, 13π/6 is not in range of cos-1 i.e., [0, π]. so, break of 13π/6 = cos-1 [cos (2π + π/6)] = cos-1 [cos π/6] = π/6
Question-2 :- Find the value of tan-1 (tan 7π/6).
Solution :-tan-1 (tan 7π/6) Here, 7π/6 is not in range of cos-1 i.e., [-π/2, π/2]. so, break of 7π/6 = tan-1 [tan (π + π/6)] = tan-1 [tan π/6] = π/6
Question-3 :- Prove that : 2sin-1 (3/5) = tan-1 (24/7).
Solution :-2sin-1 (3/5) = tan-1 (24/7) By taking L.H.S 2sin-1 (3/5) [2sin-1 x = sin-1(2x √1 - x²)] = sin-1 (2 x 3/5 √1 - (3/5)²) = sin-1 (6/5 √1 - 9/25) = sin-1 (6/5 √16/25) = sin-1 (6/5 x 4/5) = sin-1 (24/25) Let sin-1 (24/25) = θ sin θ = 24/25 cos θ = √1 - sin²θ cos θ = √1 - (24/25)² cos θ = √1 - 576/625 cos θ = √49/625 cos θ = 7/25 Now, tan θ = sin θ/ cos θ = 24/25 x 25/7 = 24/7 tan θ = 24/7 θ = tan-1 (24/7) = R.H.S
Question-4 :- Prove that: sin-1 (8/17) + sin-1 (3/5) = tan-1 (77/36).
Solution :-sin-1 (8/17) + sin-1 (3/5) = tan-1 (77/36) L.H.S sin-1 (8/17) + sin-1 (3/5) By Using [sin-1 x + sin-1 y = sin-1(x√1 - y² + y√1 - x²)] = sin-1 (8/17 √1 - (3/5)² + 3/5 √1 - (8/17)²) = sin-1 (8/17 √1 - 9/25 + 3/5 √1 - 64/289) = sin-1 (8/17 √16/25 + 3/5 √225/289) = sin-1 (8/17 x 4/5 + 3/5 x 15/17) = sin-1 (32/85 + 9/17) = sin-1 [(32 + 45)/85] = sin-1 77/85 Let sin-1 77/85 = θ sin θ = 77/85 cos θ = √1 - sin²θ cos θ = √1 - (77/85)²θ cos θ = √1 - 5929/7225 cos θ = √1296/7225 cos θ = 36/85 Now tan θ = sin θ/cos θ tan θ = 77/85 x 85/36 = 77/36 θ = tan-1 77/36 = R.H.S
Question-5 :- Prove that: cos-1 (4/5) + cos-1 (12/13) = cos-1 (33/65).
Solution :-cos-1 (4/5) + cos-1 (12/13) = cos-1 (33/65) By taking L.H.S cos-1 (4/5) + cos-1 (12/13) By Using [cos-1 x + cos-1 y = cos-1(xy - √1 - y² √1 - x²)] = cos-1(4/5 x 12/13 - √1 - (12/13)² √1 - (4/5)²) = cos-1(48/65 - √1 - 144/169 √1 - 16/25) = cos-1(48/65 - √25/169 √9/25) = cos-1[48/65 - (5/13 x 3/5)] = cos-1[48/65 - 3/13] = cos-1[(48-15)/65] = cos-1(33/65) = R.H.S
Question-6 :- Prove that : cos-1 (12/13) + sin-1 (3/5) = sin-1 (56/65).
Solution :-cos-1 (12/13) + sin-1 (3/5) = sin-1 (56/65) By taking L.H.S cos-1 (12/13) + sin-1 (3/5) Let cos-1 (12/13) = θ cos θ = 12/13 sin θ = √1 - (12/13)² sin θ = √1 - 144/169 sin θ = √25/169 sin θ = 5/13 Now, θ = sin-1 (5/13) sin-1 (5/13) + sin-1 (3/5) By using [sin-1 x + sin-1 y = sin-1(x√1 - y² + y√1 - x²)] = sin-1 (5/13 √1 - (3/5)² + 3/5 √1 - (5/13)²) = sin-1 (5/13 √1 - 9/25 + 3/5 √1 - 25/169) = sin-1 (5/13 √16/25 + 3/5 √144/169) = sin-1 (5/13 x 4/5 + 3/5 x 12/13) = sin-1 (4/13 + 36/65) = sin-1 [(20 + 36)/65] = sin-1 (56/65) = R.H.S
Question-7 :- Prove that: tan-1 (63/16) = sin-1 (5/13) + cos-1 (3/5).
Solution :-tan-1 (63/16) = sin-1 (5/13) + cos-1 (3/5) R.H.S sin-1 (5/13) + cos-1 (3/5) Firstly, Let sin-1 5/13 = θ₁ sin θ₁ = 5/13 cos θ₁ = √1 - sin²θ₁ cos θ₁ = √1 - (5/13)² cos θ₁ = √1 - 25/169 cos θ₁ = √144/169 cos θ₁ = 12/13 Now tan θ₁ = sin θ₁/cos θ₁ tan θ₁ = 5/13 x 13/12 = 5/12 θ₁ = tan-1 5/12 Secondly, Let cos-1 (3/5) = θ₂ sin θ₂ = 3/5 sin θ₂ = √1 - cos²θ₂ sin θ₂ = √1 - (3/5)² sin θ₂ = √1 - 9/25 sin θ₂ = √16/25 sin θ₂ = 4/5 Now tan θ₂ = sin θ₂/cos θ₂ tan θ₂ = 4/5 x 5/3 = 4/3 θ₂ = tan-1 4/3 Now, θ₁ + θ₂, We have, tan (θ₁ + θ₂) = [(tan θ₁ + tan θ₂) ÷ (1 - tan θ₁ tan θ₂)] = tan-1 (5/12) + tan-1 (4/3) = tan-1[(5/12 + 4/3) ÷ (1 - 5/12 x 4/3)] = tan-1[(5 + 16)/12 ÷ (1 - 5/9)] = tan-1[21/12 ÷ 4/9)] = tan-1[7/4 x 9/4] = tan-1(63/16) = L.H.S
Question-8 :- Prove that: Prove that: tan-1 (1/5) + tan-1 (1/7) + tan-1 (1/3) + tan-1 (1/8) = π/4.
Solution :-Question-9 :- Prove that : tan-1 √x = 1/2 . cos-1 (1 - x)/(1 + x), x ∈ [0,1].
Solution :-tan-1 √x = 1/2 . cos-1 (1 - x)/(1 + x) 2tan-1 √x = cos-1 (1 - x)/(1 + x) By Taking L.H.S 2tan-1 √x We using Property, 2tan-1 x = cos-1 (1 - x²)/ (1 + x²) = cos-1 [1 - (√x)²]/ [1 + (√x)²] = cos-1 (1 - x)/(1 + x) = R.H.S
Question-10 :- Prove that:
Question-11 :- Prove that:
9
Question-12 :- Prove that : 9π/8 - 9/4 sin-1 1/3 = 9/4 sin-1(2√2)/3.
Solution :-9π/8 - 9/4 sin-1 1/3 = 9/4 sin-1(2√2)/3 Taking out common value 9/4 of both sides then, 9/4 x (π/2 - sin-1 1/3) = 9/4 sin-1(2√2)/3 π/2 - sin-1 1/3 = sin-1(2√2)/3 cos-1 1/3 = sin-1 (2√2)/3 By taking of L.H.S let cos-1 1/3 = θ cos θ = 1/3 sin θ = √1 - cos²θ sin θ = √1 - (1/3)² sin θ = √1 - 1/9 sin θ = √8/9 sin θ = (2√2)/3 θ = sin-1(2√2)/3 = R.H.S
Question-13 :- Solve: 2tan-1(cos x) = tan-1(2 cosec x)
Solution :-
2tan-1(cos x) = tan-1(2 cosec x)
tan-1[(2 cos x) ÷ (1 - cos²x)] = tan-1(2 cosec x)
(2 cos x) ÷ (1 - cos²x) = 2 cosec x
2 cos x ÷ sin²x = 2 ÷ sin x
cos x = sin x
tan x = 1
tan x = tan π/4
x = π/4
Question-14 :- Solve: tan-1[(1 - x) ÷ (1 + x)] = 1/2 tan-1x, (x > 0).
Solution :-tan-1[(1 - x) ÷ (1 + x)] = 1/2 tan-1x tan-1 1 - tan-1 x = 1/2 tan-1x tan-1 (tan π/4) = 1/2 tan-1x + tan-1 x π/4 = 3/2 tan-1x tan-1x = π/6 x = tan (π/6) x = 1/√3
Question-15 :- sin(tan-1x), |x|<1 is equal to
sin(tan-1x) Let tan-1x = θ x = tan θ sin θ = x/(√1 + x²) θ = sin-1[x/(√1 + x²)] tan-1x = sin-1[x/(√1 + x²)] sin(tan-1x) = sin(sin-1[x/(√1 + x²)]) sin(tan-1x) = x/(√1 + x²) Option is D.
Question-16 :- sin-1(1 - x) - 2 sin-1x = π/2, then x is equal to
Question-17 :-
