Solution of tan-12x + tan-1 3 x = π/4
-1 or 1/6
1 or -1/6
-1 or 6
1 or -6
tan (735o) = tan (2 × 360o + 15o)
Sin 35o
Cos 15o
tan 15o
cot 15o
Sin (1/2 cos-1 4/5)
The value of is
3/16
3/8
3/4
1/2
2
1
√2
The value of tan 40o + tan 20o + √3 tan 20o tan 40o is equal to
√12
1/√3
√3
√3/2
Sin-1 (1/x) =
sin x
cosec-1x
cos x
cos (1/x)
Area of Δ ABC is
1/2 ab cos C
1/2 ab sin c
1/2 ab cos B
1/2 ca sin B
ΔABC is right angled at C, then tan A + tan B is equal to.
a + b
If sin θ = 3/8 and θ is acute, find sin 2 θ
