1/cos θ =
sin θ
cosec θ
sec θ
tan θ
If cosec θ = √2, then cot θ =
0
1
√2 + 1
√2 - 1
If 2 sin θ = √3, then cos θ =
√3/2
2/√3
1/2
2
cosec 74 + tan 81 equal to
sec18 + cot 10
sec17+ cot 11
sec16 + cot 9
sec19+ cot 12
In the figure find sin (90o - θ )
4/5
3/5
5/3
5/4
In right triangle Δ ABC, ∠ B = 90o, tan C = 1/2. If AC = 6 cm, then AB =
6/√5 cm
2 cm
3/√5 cm
√5 cm
If cosec θ = √5, then cot θ - cos θ =
If sec θ = √3, then sin θ + cos θ =
√2
√6 + √3
If cos θ = 3/4, then sin θ =
3/4
4/3
√7/4
√2/3
cot θ =
cos θ/ sinθ
1/cos θ
sin θ/cosθ
1/sec θ
