In the adjoining figure, Sinθ = . Then BC =
85 m
65 m
95 m
15 m
(Cosec A - Sin A) (Sec A - Cos A) (tan A + Cot A) =
0
1
Sin A
Cos A
In the adjoining figure
45°
30°
60°
50°
If Cos θ = , then Sin θ is equal
Cos4x - Sin4x =
2Sin2x - 1
2Cos2x - 1
1+ 2sin2x
1 - 2Cos2x
(1- Sin2θ) Sec2θ
tan2θ
Cos2θ
9 tan2θ - 9 Sec2θ =
9
-9
Sin (90 - θ) Cos θ + Cos (90 - θ) Sinθ =
2
-1
(1 - Cos2θ) (1 + Cot2θ)
Sin2θ
