Answer:
They are both parallel lines
Step-by-step explanation:
they both have the slope of -5/4
Thus L.H.S = R.H.S that is 2/√3cosx + sinx = sec(Π/6-x) is proved
We have to prove that
2/√3cosx + sinx = sec(Π/6-x)
To prove this we will solve the right-hand side of the equation which is
R.H.S = sec(Π/6-x)
= 1/cos(Π/6-x)
[As secƟ = 1/cosƟ)
= 1/[cos Π/6cosx + sin Π/6sinx]
[As cos (X-Y) = cosXcosY + sinXsinY , which is a trigonometry identity where X = Π/6 and Y = x]
= 1/[√3/2cosx + 1/2sinx]
= 1/(√3cosx + sinx]/2
= 2/√3cosx + sinx
R.H.S = L.H.S
Hence 2/√3cosx + sinx = sec(Π/6-x) is proved
Learn more about trigonometry here : brainly.com/question/7331447
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Answer:
5k-17
Step-by-step explanation:
-14+5k-3=5k-14-3=5k-17
The surface area of a right cylinder is the sum of the areas of the lateral face and the circular bases. The lateral face is really just a rectangle with the same height as the cylinder and a length equal to the circumference of the circular base.
If the radius of the base is

and the height is

, then the surface area of the cylinder is

You're given that the area is

and the radius is

, so you have