Question

If 5secθ = 13 find the value of 2sinθ+3tanθ 4sinθ−9cosθ​

Answer 1

Step-by-step explanation:

given sec0=13/5=h/b

h=13 and b=5

p√(13)^2-(5)^2=√(169-25)=√144=12

find

2sin0+3tan0

2×p/h+3×p/b

2×12/13+3×12/5

24/13+36/5=(120+468)/65=588/65

4sin0-9cos0

4×12/13-9×5/13

48/13-45/13

3/13