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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For both part A and part B, the cross-section would be a square.
For part c, the reasons the shapes are the same is the 3-D figure of a cube has 6 faces that are all squares. Since every face is a square, you cannot cut a cross-section that is any shape but a square. Cross-sections in 3-D prisms can reflect the face they are nearest to, so with every face being identical on a cube, the cross sections are also identical.
I hope this helps! :)
Answer:
The answer is C) Y = 4
Step-by-step explanation:
In A, the slope is defined as 2.
In B, the slope is defined as 1/2.
In D, the slope is undefined.
Answer:
10 ft
Step-by-step explanation:
perpendicular (p)= 6 ft
base(b) =8 ft
hypotenuse (h) = ?
we know by using Pythagoras theorem
h = 
= 
= 10 ft
hope it helps :)
Answer:
Step-by-step explanation:
I have answered ur question
