Question

If 3sinA=4cosA
Find:
1) sinA
2)cosA
3)tan²A-sec²A

Answer 1

3sinA =4cosA
SinA /cosA =4/3
tanA =4/3
using Pythagoras theorem,
hypotenuse =5
perpendicular=4
base=3
1)sinA =P/H
=4/5
2)cosA=B/H
=3/5
3)tan^2A=(4/3)^2
=16/9
sec^2A=(5/3)^2
=25/9
.˙.tan^2A-sec^2A=(16/9)-(25/9)
=(-9/9)
=-1

Answer 2

3sinA = 4cosA
4/3= sinA/cosA
4/3 = tanA
e.g tanA = 4/3 = P/b
h = √(p² + b²) = 5

(1) sinA = P/h = 4/5
(2) cosA = b/h = 3/5
(3) tan²A - sec²A
= - { sec²A - tan²A }
= -1