Tan x /(1 +sec x) + (1+sec x) /tan x
Tan x=sin x / cos x
1+ sec x=1 +1/cos x=(cos x+1)/cos x
Therefore:
tan x /(1 +sec x) =(sin x/cos x)/(cos x+1)/cos x=
=(sin x * cos x) / [cos x* (cos x+1)]=sin x /(Cos x+1)
(1+sec x) /tan x=[(cos x+1)/cos x] / (sin x/cos x)=
=[cos x(cos x+1)]/(sin x *cos x)=(cos x+1)/sin x
tan x /(1 +sec x) + (1+sec x) /tan x=
=sin x /(Cos x+1) + (cos x+1)/sin x=
=(sin²x+cos²x+2 cos x+1) / [sin x(cos x+1)]=
Remember: sin²x+cos²x=1⇒ sin²x=1-cos²x
=(1-cos²x+cos²x+2 cos x+1) / [sin x(cos x+1)]=
=2 cos x+2 / [sin x(cos x+1)]=
=2(cos x+1) / [sin x(cos x+1)]=
=2 /sin x
Answer : tan x /(1 +sec x) + (1+sec x) /tan x= 2/sin x
Set the whole expression = to 0 and solve for x.
3x^(5/3) - 4x^(7/3) = 0. Factor out x^(5/3): x^(5/3) [3 - 4x^(2/3)] = 0
Then either x^(5/3) = 0, or 3 - 4x^(2/3) = 0.
In the latter case, 4x^(2/3) = 3.
To solve this: mult. both sides by x^(-2/3). Then we have
4x^(2/3)x^(-2/3) = 3x^(-2/3), or 4 = 3x^(-2/3). It'd be easier to work with this if we rewrote it as
4 3
--- = --------------------
1 x^(+2/3)
Then
4
--- = x^(-2/3). Then, x^(2/3) = (3/4), and x = (3/4)^(3/2). According to my 3 calculator, that comes out to x = 0.65 (approx.)
Check this result! subst. 0.65 for x in the given equation. Is the equation then true?
My method here was a bit roundabout, and longer than it should have been. Can you think of a more elegant (and shorter) solution?
12 is divisible by 1,2,3,4,6,12 and 27 is by 1,3,9,27 so the GCF is 3
U+xy
On replacing for u, x and y.
=9+9*7
=9+63
=72
