In Δ PQR,∠ Q = 90o, PQ = 4 cm and PR = 9 cm, then the value of sinP is
10/9
√95/9
√65/9
√17/4
If sec θ = 2 then sin θ =
1/2
3/2
1/√3
√3/2
sin θ × _______ = 1
cos θ
cosec θ
sec θ
cot θ
cot θ =
cos θ/ sinθ
1/cos θ
sin θ/cosθ
1/sec θ
sin2 50+ cos2 50 =
1
0
1.25
2.32
In the figure find sin (90o - θ )
4/5
3/5
5/3
5/4
If cosec θ = √5, then cot θ - cos θ =
If cosec θ = √2, then cot θ =
√2 + 1
√2 - 1
If sec θ = √3, then sin θ + cos θ =
√2
√6 + √3
If a sin θ = 1 and b tan θ = 1, then the relation between a and b is
a2 = b2
a2 - b2 = 1
a2 + b2 = 1
a2 + b2 = -1
