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
Sin (cos-1 (3/5) =
1/5
3/4
4/5
1/3
The value of is
0
2
3
1
The product s(s-a) (s-b) (s-c) is equal to
Δ
Δ2
2 Δ
Δ/s
Solution of tan-12x + tan-1 3 x = π/4
-1 or 1/6
1 or -1/6
-1 or 6
1 or -6
cos 15o =
Sin-1 (1/x) =
sin x
cosec-1x
cos x
cos (1/x)
Sin (1/2 cos-1 4/5)
Cos (tan-1 3/4) =
2/5
1/4
In a ΔABC, ∠A = π/2, then cos2 B + cos2 C equals
-2
-1
