In right triangle Δ ABC, ∠ B = 90^{o}, tan C = ^{1}/2. If AC = 6 cm, then AB =
^{6}/√5 cm
2 cm
^{3}/√5 cm
√5 cm
If cosec θ = √2, then cot θ =
0
1
√2 + 1
√2 - 1
^{1}/cos θ =
sin θ
cosec θ
sec θ
tan θ
If cos θ = ^{3}/4, then sin θ =
^{3}/4
^{4}/3
^{√7}/4
^{√2}/3
If cosec θ = √5, then cot θ - cos θ =
^{1}/5
^{1}/3
^{3}/5
^{5}/4
If 2 sin θ = √3, then cos θ =
^{√3}/2
^{2}/√3
^{1}/2
2
If a sin θ = 1 and b tan θ = 1, then the relation between a and b is
a^{2} = b^{2}
a^{2} - b^{2} = 1
a^{2} + b^{2} = 1
a^{2} + b^{2} = -1
sin^{2} 50+ cos^{2} 50 =
1.25
2.32
cot θ =
^{cos θ}/ sinθ
^{1}/cos θ
^{sin θ}/cosθ
^{1}/sec θ