How do I convert the left side to be equal with the right,,,,,
2006-12-18 14:10:26 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
in triangle ABC, with right angle at C, if c=6 and a=4, then cos A=
2006-12-18 15:16:14 · update #1
CSC = 1/Sin
Cot = Cos/Sin
then CSC/COT = 1/Sin/Cos/Sin
= 1/Sin * Sin/Cos
= 1/Cos
= Sec
thats it
2006-12-18 14:37:38 · answer #1 · answered by Ramanadhan C 2 · 0⤊ 0⤋
(csc(x)/cot(x)) = sec(x)
(1/sin(x))/(cos(x)/sin(x)) = sec(x)
(1/sin(x)) * (sin(x)/cos(x)) = sec(x)
(sin(x))/(sin(x)cos(x)) = sec(x)
1/cos(x) = sec(x)
sec(x) = sec(x)
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in triangle ABC, with right angle at C, if c=6 and a=4, then cos A=
a^2 + b^2 = c^2
4^2 + b^2 = 6^2
16 + b^2 = 36
b^2 = 20
b = sqrt(20)
Using this, you get
cos(A) = (2sqrt(5)/6)
cos(A) = (sqrt(5))/3
cos(A) = about .7454
2006-12-18 23:30:31 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
csc/cot =
(1/sin)/(cos/sin) =
1/cos =
sec
2006-12-18 22:12:41 · answer #3 · answered by Philo 7 · 0⤊ 0⤋
csc=1/sin
cot=cos/sin
csc/cot=(1/sin)(sin/cos)
=1/cos
=sec
2006-12-18 22:13:47 · answer #4 · answered by raj 7 · 0⤊ 0⤋
Make everything into sins and coss.
(1/sinx)/(cosx/sinx)=1/cosx
1/cosx=1/cosx
2006-12-18 22:13:36 · answer #5 · answered by knock knock 3 · 0⤊ 0⤋