Trigonometric Ratios for Sum of Two Angles
Trending Questions
- (n−1)24n
- (n+1)24n
- (n−1)24n2
- (n+1)24n2
sin−1[x√1−x−√x√1−x2]=
sin−1x−sin−1√1−x
sin−1x+sin−1√1−x
sin−1x−sin−1√x
None
1tan3A−tanA - 1cot3A−cotA =
tanA
tan2A
cotA
cot2A
- tan−112
- tan−12+tan−13
- tan−12
- tan−112+cot−112
If sin−113+sin−123=sin−1 x, then x is equal to
[Roorkee 1995]
- 0
- √5−4√29
√5+4√29
x2
- 16x2–12x+1=0
- x2–12x+1=0
- 16x2–2x–1=0
- None
If cos3θ = αcosθ+βcos3θ, then (α, β) =
(3, 4)
(4, 3)
(-3, 4)
(3, -4)
Which of the following is the CORRECT combination ?
- a→p, b→q, c→r
- a→r, b→q, c→p
- a→q, b→r, c→p
- a→r, b→p, c→q
- 2116
- 6316
- 6352
- 3352
If cos3θ = αcosθ+βcos3θ, then (α, β) =
(3, 4)
(4, 3)
(-3, 4)
(3, -4)
- 833
- 733
- 3356
- 5633
If cos(θ−α) = a, sin(θ−β) = b,
then cos2(α−β) + 2ab sin(α−β) is equal to
4a2b2
a2−b2
a2+b2
-a2b2
- cot(θ+ϕ)=6365
- tan(θ+ϕ)=5633
- sin(θ+ϕ)=5665
- cos(θ−ϕ)=6365
- True
- False
If sin A = 45 and cos B = - 1213, where A and B lie in first
and third quadrant respectively, then cos(A + B) =
5665
- 5665
1665
- 1665
- 16x2–12x+1=0
- x2–12x+1=0
- 16x2–2x–1=0
- None
- 3−√3
- 2(3−√3)
- 2(√3−1)
- 2(2−√3)
- [3, 4]
- [−3, 4]
- [−4, −3]
- [−4, 3]
- 1
- 2
- 3
- √3
- tan15x+tan11x+tan4x=tan15xtan11xtan4x
- tan15x−tan11x−tan4x=tan15xtan11xtan4x
- cot2xcot4x−cot4xcot6x−cot2xcot6x=1
- cot4xcot6x−cot2xcot4x−cot2xcot6x=1
- 60∘
- 30∘
- 45∘
- 90∘
- f is one-one function
- Range of f is{2, 3}
- f is many-one function
- Range of f is {0, 1}
If tan A = - 12 and tan B = - 13, then A + B =
[IIT 1967; MNR 1987; MP PET 1989]
π4
3π4
5π4
None of these
Column-IColumn-II (A)f1(π2)(P)0(B)f3(3π16)(Q)√2−1(C)f4(2π11)(R)√2+1(D)f5(π28)(S)1(T)2−√3
- (A)→(P)(B)→(Q)(C)→(R)(D)→(S)
- (A)→(S)(B)→(Q)(C)→(P)(D)→(R)
- (A)→(S)(B)→(Q)(C)→(P)(D)→(T)
- (A)→(T)(B)→(R)(C)→(P)(D)→(R)
- tan(θ)−tan(θ16)
- tan(θ16)−tan(θ)
- tan(θ)−tan(θ32)
- tan(θ2)−tan(θ64)
If tanθ = sinα−cosαsinα+cosα, then sinα+cosα and
sinα−cosα must be equal to
√2cosθ, √2sinθ
√2sinθ, √2cosθ
√2sinθ, √2sinθ
√2cosθ, √2cosθ
Column-IColumn-II (A)f1(π2)(P)0(B)f3(3π16)(Q)√2−1(C)f4(2π11)(R)√2+1(D)f5(π28)(S)1(T)2−√3
- (A)→(P)(B)→(Q)(C)→(R)(D)→(S)
- (A)→(S)(B)→(Q)(C)→(P)(D)→(R)
- (A)→(S)(B)→(Q)(C)→(P)(D)→(T)
- (A)→(T)(B)→(R)(C)→(P)(D)→(R)