Which choice is the equation of the line that passes through the point (5, 19) and has a slope of
1 answer:
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Take derivitive
note
the derivitive of sec(x)=sec(x)tan(x)
so
remember the quotient rule
the derivitive of
so
the derivitive of
so now evaluate when t=pi
we get
sec(pi)=-1
tan(pi)=0
we get
slope=1/pi
use slope point form
for
slope=m and point is (x1,y1)
equation is
y-y1=m(x-x1)
slope is 1/pi
point is (pi,1/pi)
y-1/π=1/π(x-π)
times both sides by π
πy-1=x-π
πy=x-π+1
y=(1/π)x-1+(1/π)
or, alternately
-(1/π)x+y=(1/π)-1
x-πy=π-1
note
the derivitive of sec(x)=sec(x)tan(x)
so
remember the quotient rule
the derivitive of
so
the derivitive of
so now evaluate when t=pi
we get
sec(pi)=-1
tan(pi)=0
we get
slope=1/pi
use slope point form
for
slope=m and point is (x1,y1)
equation is
y-y1=m(x-x1)
slope is 1/pi
point is (pi,1/pi)
y-1/π=1/π(x-π)
times both sides by π
πy-1=x-π
πy=x-π+1
y=(1/π)x-1+(1/π)
or, alternately
-(1/π)x+y=(1/π)-1
x-πy=π-1
Answer:
Cos θ = -4/5
Tan θ = -3/4
Step-by-step explanation:
The question is as following:
Select the correct answer from each drop-down menu.
Angle θ lies in the second quadrant, and sin θ =3/5.
cos θ = -4/5, -3/5, 3/5, 4/5
tan θ = -4/3, -3/4, 3/4, 4/3
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Since, the angle lies in second quadrant (negative x axis and positive y axis) we can deduce that cos θ is negative and tan θ is also negative.
If sin θ =3/5
∴ Cos θ = -4/5 ⇒ because θ lies in the second quadrant.
And
Tan θ = (sin θ)/(cos θ) = (3/5) / (-4/5) = -3/4.